pH - AP Chemistry

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Question

What is the pH of a soution containing .0001 M HCl?

Answer

The pH of a solution is determined by taking the negative log of the concentration of hydrogen ions in solution. HCl is strong acid so it completely dissociates in solution. So adding .0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log\[.0001\] =4, so the pH of the solution =4.

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Question

What is the pOH of a solution that has a \[H+\] = 3.2 X 10-7 mol?

Answer

\[H+\] = 3.2 X 10-7 mol

pH = -log \[H+\]

= 7 - log 3.2

= 7 -.505

=6.495

pOH = 14 - pH

= 7.5

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Question

Which of the following solutions contains the greatest number of H+ ions?

Answer

This question asks for the solution with the greatest number H+ ions, we can also approach this problem as if we are looking for the solution with the lowest pH. HCl is strong acid, therefore it will dissociate completely in solution. So the solution containing 0.010 mL of HCL contains 0.010 mL of H+ ions in solution, in addition to the H+ ions that are already in solution due to the auto-ionization of water. The only other solution that could have a pH less than 7 would be the one with 0.010 mL of CH3OH in excess water, becasue CH3OH is very slightly acidic. But since it is compared with an equal volume of HCl which is strong acid, it can be said that the most H+ ions will be found in the solution containing a small amount of strong acid, HCl.

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Question

What is the approximate pH of a 1.0 M solution of soluble CaCO3?

Answer

Calcium carbonate is a base since it's the salt of a strong base (NaOH) and a weak acid (carbonic acid). The only basic pH on the list is 8, making it the correct answer.

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Question

What is the pH of a 1 * 10–3M solution of H2CO3 acid? (pKa is 6.4)

Answer

The –log of the pKa will give you the Ka, so take the –log (6.4), which gives you approximately 4 * 10–7. The Ka expression is set up with products over reactants (hydrogen carbonate ion * hydrogen ion/carbonic acid). The undissociated carbonic acid is 0.001M, and you should use the variable 'x' to account for how much it dissociates and how many of the ions are produced. Ka = 4 * 10–7 = x2/0.001 ends up being your Ka expression, if you assume x is negligible compared to the original concentration of 0.001. Solving for x, you get 2 * 10–5. This is the hydrogen ion concentration. pH = –log (hydrogen ion concentration), so pH = –long(2 * 10–5), which is approximately 4.7.

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Question

If the Ksp of Mg(OH)2 is 1.2 * 10–11 and the magnesium ion concentration is 1.2 * 10–5M, at what pH does the Mg(OH)2 compound begin to precipitate?

Answer

The expression for Ksp is Ksp = \[Mg2+\]\[OH–\]2.

Thus, \[OH–\] = √(1.2 * 10–11)/(1.2 * 10–5)

\[OH–\] = 1 * 10–3

Thus, pH = –log(1 X 10–3) = 3

pH = 14 – 3 = 11

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Question

What is the range of possible hydrogen ion concentrations in acidic solution?

Answer

For a solution to be acidic, it must have a pH between 1 and 6.99, since 7 is neutral; pH is –log(hydrogen ion concentration), the range of possible values is between 10–1 and 10–6.999.

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Question

What is the range of possible hydrogen ion concentrations in basic solution?

Answer

For a solution to be basic, it must have a pH between 8 and 14, since 7 is neutral; pH is –log(hydrogen ion concentration); the range of possible values is between 10–13 and 10–7.001.

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Question

What is the pH of a 0.05M solution of hydroflouric acid?

Answer

The Ka = \[H+\]\[F–\]/\[HF\]. The beginning concentration of HF is given as 0.05, and we can use x as the variable that accounts for how much of the HF is lost, along with how much of the H+ and F– are formed.

Thus, Ka = _x_2/.05 if you use the approximation that x is negligible compared to the starting concentration of HF.

Solving for x, x = 1 * 10–3. This is the H+ ion concentration.

pH = –log(H+) = –log(1.0 * 10–3)) = 3

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Question

A chemist adds 625g of solid $NaOH$ to 500mL of 16M $H_2SO_4$ . What is the pH of the solution after it reaches equilibrium?

Answer

Consider the reaction of $NaOH$ and $H_2SO_4$ :

$2NaOH+H_2SO_4\rightarrow Na_2SO_4\hspace{1 mm}+2H_2O$

Now we will calculate the moles of $H_2SO_4$ in the solution prior to adding base.

$500\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{16\hspace{1 mm}moles\hspace{1 mm}H_2SO_4}{1\hspace{1 mm}L}=8.00\hspace{1 mm}moles\hspace{1 mm}H_2SO_4$

We will then calculate the amount of moles of $H_2SO_4$ that react with the base.

$625\hspace{1 mm}g\hspace{1 mm}NaOH\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaOH}{40\hspace{1 mm}g\hspace{1 mm}NaOH}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}H_2SO_4}{2\hspace{1 mm}moles\hspace{1 mm}NaOH}=7.81\hspace{1 mm}moles\hspace{1 mm}H_2SO_4$

We will then calculate the remaining moles of $H_2SO_4$ :

$8.00\hspace{1 mm}moles\hspace{1 mm}H_2SO_4 -\hspace{1 mm}7.81\hspace{1 mm}moles\hspace{1 mm}H_2SO_4=0.19\hspace{1 mm}moles\hspace{1 mm}H_2SO_4$

We will then calculate the new concentration of sulfuric acid:

$\frac{0.19\hspace{1 mm}moles\hspace{1 mm}H_2SO_4}{0.5\hspace{1 mm}L}=0.38\hspace{1 mm}M$

Sulfuric acid is a diprotic acid, so the hydrogen ion concentration is 0.76 M.

$pH=-log[H^{+}]=-log(0.76)=0.12$

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Question

A chemist adds $4.2\hspace{1 mm}g\hspace{1 mm}KOH$ to mL of water. What is the pH of the solution?

Answer

First, we will calculate $[KOH]$ as follows:

$\frac{4.2\hspace{1 mm}g\hspace{1 mm}KOH}{500\hspace{1 mm}mL}\times\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}KOH}{56.1\hspace{1 mm}g\hspace{1 mm}KOH}=1.497\times10^{-1}\hspace{1 mm}M$

Now we will calculate the pOH:

$pOH=-log[OH^-]$

In solution $KOH_{(s)}\rightarrow K^{+}{(aq)}+OH^{-}{(aq)}$ , so $[KOH]=[OH^-]$ .

$pOH=-log[1.497\times 10^{-1}]=8.248\times 10^{-1}$

$pH=14-pOH=14-0.8248=13.18$

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Question

A chemist boils off the water from 627 mL of a solution with a pH of 10.2. The only dissolved compound in the solution is $NaOH$ . How much solid remains?

Answer

The pH of the solution is 10.2, and we know:

$pH\hspace{1 mm}+\hspace{1 mm}pOH=14$

$pOH=14-pH=14-10.2=3.8$

$pOH=-log[OH^-]$

$3.8=-log[OH^-]$

$-3.8=log[OH^-]$

$[OH^-]=10^{-3.8}=1.58\times 10^{-4}$

And since $NaOH_{(s)}\rightarrow Na^+{(aq)}+OH^-{(aq)}$ , $[NaOH]=1.58\times 10^{-4}$ .

$627\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{1.58\times 10^{-4}\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=4.03\times 10^{-3}\hspace{1 mm}g\hspace{1 mm}NaOH$

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Question

A chemist adds 10mL of a solution with a pH of 3.00 to 50mL of water. What is the pH of the new solution?

Answer

First, we will determine the amount of hydronium ions in the first solution. This allows us to measure the moles of hydronium ion that are transferred to water.

$[H+]=10^{-3}=0.00100\hspace{1 mm}M$

$10\hspace{1 mm}mL\times\frac{1\hspace{ 1 mm}L}{1000\hspace{ 1 mm}mL}\times\frac{0.001\hspace{ 1 mm}mol}{1\hspace{ 1 mm}L}=1.00\times10^{-5}\hspace{ 1 mm}mol$

Now we can calculate the hydronium ion concentration in the new solution, using the amount transferred and the volume of water, and can solve for pH.

$\frac{1.0\times10^{-5}mol}{60\hspace{ 1 mm}mL}\times\frac{1000\hspace{ 1 mm}mL}{1\hspace{ 1 mm}L}=1.67\times 10^{-4}\hspace{ 1 mm}M$

$pH=-log[1.67\times 10^{-4}]=3.78$

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Question

A chemist mixes of a $1.0M\ NaOH$ solution with of a $1.0M\ Ba(OH)_2$ solution. Assuming the two solutions are additive, what is the pH of the resulting solution?

Answer

To find the pH of the solution after mixing the individual components, one first has to determine the final hydroxide concentration, \[-OH\].

$[OH^-]=\frac{mol OH^-}{total volume}$

(Remember that 1 mole of Ba(OH)2 will give 2 moles of \[-OH\]).

$[OH^-]=\frac{0.150mol}{0.10L}=1.5M$

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Question

Which two solutions, when combined in equal amounts, will produce the lowest pH?

Answer

HCl is a strong acid, while NaCl is a salt. Mixing these two will result in a pH < 1.

As for the other choices, mixing 2.0M HCl and 2.0M KOH will result in the formation of KCl and H2O, resulting in a neutral pH. Acetic acid (pKa= 4.76) is a weak acid, so it will have a pH less than 7 but greater than 1. Finally, NH3 and NH4+ is a buffer system between a weak base and weak acid, and if equal amounts are present, the pH will be the same as the pKa of NH4+, which is around 9.

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Question

1mol of hydrobromic acid (HBr) is added to water, resulting in 1L of solution. Worried that the acid is too strong, a professor adds more water, increasing the volume of solution to 2L.

What was the pH of the solution before adding water? What was the pH after adding the water?

Answer

The pH of a solution is given by the equation .

We can assume that hydrobromic acid, as a strong acid, will dissociate completely. As a result, there will be 1mol of protons in the initial solution. Molarity is a method of measuring the concentration of an agent in a solution and is defined by the equation below.

$M=\frac{mol\ solute}{L\ solution}$

Since we have 1mol of protons and 1L of solution, the molarity of protons is simply 1M in the initial solution (1mol/1L). We can plug this into the pH equation and solve.

After the dilution with water, the concentration of protons will change. Since the volume of the solution has doubled, the concentration of protons has been halved.

$M=\frac{1mol}{2L}=0.5M$

* Key points to remember!

The lower the pH, the more acidic the solution. This is why the diluted solution was 0.3 and the original, more acidic solution was 0.

pH is a logarithmic scale. As a result, a pH of 2 is ten times more acidic than a pH of 3. This is why halving the concentration DID NOT halve the pH.

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Question

What is the pH of a solution if you dilute 10mL of a 0.56M NaOH solution into 100mL of water?

Answer

We first need to calculate the moles of NaOH in 10mL of a 0.56M solution.

$10\hspace{1 mm}mL\cdot\frac{0.56\hspace{1 mm}mol\hspace{1 mm}NaOH}{1000\hspace{1 mm}mL}=5.6\cdot 10^{-3}\hspace{1 mm}mol\hspace{1 mm} NaOH$

Now, in the new volume, the concentration is equal to this same number of moles divided by the new volume. The new volume is the sum of the two solutes together, so the sum of 10mL of solution and 100mL of water.

$\frac{5.6\cdot 10^{-3}\hspace{1 mm}mol\hspace{1 mm}NaOH}{110\hspace{1 mm}mL}\cdot\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}= 5.09\cdot 10^{-2}\hspace{1 mm}M$

Now we need to calculate pH. Since NaOH will dissociate one hydroxide ion for every molecule of NaOH in solution, the concentration of hydroxide is equal to the concentration of NaOH.

$[NaOH]=[OH^-]=5.09\cdot 10^{-2}M$

$pOH=-log[OH^-]=-log[5.09\cdot 10^{-2}]=1.29$

Knowing the pH allows us to solve for the pH.

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Question

A solution has a hydrogen ion concentration of $2.5 * 10^{-6}$ . What is the pH of this solution?

Answer

pH can be found by taking the negative log of the hydrogen ion concentration.

$[H^+]=2.510^{-6}$

$pH = -log(2.5 10^{-6})$
We may not know the exact value of this calculation, but we know that it will be somewhere between 5 and 6.

$10^{-5}>2.510^{-6}>10^{-6}$

$-log(10^-5)<-log(2.510^{-6})<-log(10^{-6})$

$5<-log(2.5*10^{-6})<6$

Only one answer choice fits this range. The answer is about 5.6.

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Question

Find the pH of a 0.03M solution of nitric acid.

Answer

Nitric acid is a strong acid, meaning that it will dissociate completely in water.

$HNO_3\rightleftharpoons H^++NO_3^-$

As a result, a 0.03M nitric acid solution will dissociate completely, resulting in a proton concentration of 0.03M.

$0.30M\ NO_3*\frac{1mol\ H^+}{1mol\ HNO_3}=0.30M\ H^+$

Now that we know the proton concentration, we can use the equation for pH to find the pH value of the solution.

$\small pH = -log[H^{+}]$

$\small pH = -log[0.03]$

$\small pH = 1.52$

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Question

Find the pH of a 0.1M solution of hydrogen cyanide.

$\small K_{a} = 6.2*10^{-10}$

Answer

Hydrogen cyanide will dissociate according to the following reaction.

$\small HCN_{(aq)} \rightarrow H^{+}_{(aq)} + CN^{-}_{(aq)}$

Hydrogen cyanide is a weak acid, meaning that it will not dissociate completely in solution. We need to use the acid dissociation constant in order to determine the concentration of protons in solution at equilibrium. We can solve for this value using an ICE table.

I. The initial concentration for the acid of 0.1M. There are no protons or cyanide ions in the solution initially, so they both have concentrations of 0.0M.

C. The concentration of both ions will increase by an unknown amount, $\small x$ , while the acid concentration will decrease by a concentration of $\small x$ .

E. Write the equilibrium expression and compare it to the dissociation constant.

$\small K_{a} = \frac{[H^{+}][CN^{-}]}{[HCN]} = 6.210^{-10}$

$\small \frac{[x][x]}{[0.1-x]} = 6.210^{-10}$

The value of $\small x$ will be small enough that it is negligible in comparison to 0.1, and can be ignored in the denominator.

$\frac{[x][x]}{0.1}=6.210^{-10}$

$x^2=6.210^{-11}$

$\small x = 7.8710^{-6}$

From our set up, we know that the value of $\small x$ is equivalent to the concentration of hydrogen ions at equilibrium. Using this value, we can calculate the pH of the solution.

$pH=-\log{[H^+]}$

$pH=-\log{(7.8710^{-6})}$

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