pH - AP Chemistry

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Question

Which of the following techniques will decrease the pH of a solution?

Answer

Increasing the concentration of protons of a solution will make the solution more acidic; therefore, it lowers the solution’s pH. Decreasing the concentration of protons will make the solution more basic, raising the pH. Adding more acid of the same molarity of the original solution will not increase the concentration of protons and will not increase acidity or lower the pH. Increasing the amount of hydroxide ions will make the solution more basic and raise the pH. Increasing the amount of solvent will lower the concentration, affecting molarity and not lowering the pH.

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Question

Determine the pH of a solution that is $0.40M \text{HCl}$ .

Answer

Since $\text{HCl}$ is a strong acid, the concentration of is equal to the concentration of the acid itself.

Thus, .

Recall how to find the pH of a solution:

$\text{pH}=-\log([H_3O^+])$

Plug in the given hydronium ion concentration to find the pH of the given solution.

$\text{pH}=-\log(0.40)=0.40$

Remember to maintain the correct number of significant figures.

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Question

Find the pH for a solution that is by mass. Assume a density of $1.01\frac{g}{mL}$ for the solution.

Answer

Start by assuming that there is liter of the solution. From this, we can use the given density to find the mass of the solution.

$\text{Mass of solution}=1000\text{mL of Solution} \cdot \frac{1.01g}{mL}=1010\text{g of solution}$

Next, find the mass of $\text{HBr}$ that is present in the solution.

$\text{Mass of HBr}=0.00520(1010g)=5.252\text{g of HBr}$

Now, find the number of moles of $\text{HBr}$ that is present in the solution.

$5.252\text{g HBr}\frac{1\text{mol HBr}}{80.91\text{g HBr}}=0.06491\text{moles of HBr}$

Since we initially assumed that we had liter of the solution, we now also know the concentration of $\text{HBr}$ in this solution.

$[\text{HBr}]=0.06491M$

Since $\text{HBr}$ is a strong acid, the concentration of hydronium ions in the solution will be the same as the concentration of $\text{HBr}$ .

$[\text{HBr}]=[H_3O^+]=0.06491M$

Recall how to find the pH of a solution.

$\text{pH}=-\log([H_3O^+])$

$\text{pH}=-\log(0.06491)=1.188$

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Question

Find the pH of a solution that is $5.06% \text{HI}$ by mass. Assume a density of $1.01\frac{g}{mL}$ for the solution.

Answer

Start by assuming that there is liter of the solution. From this, we can use the given density to find the mass of the solution.

$\text{Mass of solution}=1000\text{mL of Solution} \cdot \frac{1.01g}{mL}=1010\text{g of solution}$

Next, find the mass of $\text{HI}$ that is present in the solution.

$\text{Mass of HI}=0.0506(1010g)=51.106\text{g of HI}$

Now, find the number of moles of $\text{HI}$ that is present in the solution.

$51.106\text{g HI}\frac{1\text{mol HI}}{127.91\text{g HI}}=0.3995\text{ moles of HI}$

Since we initially assumed that we had liter of the solution, we now also know the concentration of $\text{HI}$ in this solution.

$[\text{HI}]=0.3995M$

Since $\text{HI}$ is a strong acid, the concentration of hydronium ions in the solution will be the same as the concentration of $\text{HI}$ .

$[\text{HI}]=[H_3O^+]=0.3995M$

Recall how to find the pH of a solution.

$\text{pH}=-\log([H_3O^+])$

$\text{pH}=-\log(0.3995)=0.398$

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Question

What is the pH of a solution that has \[OH-\] 1 X 10–4 M?

Answer

pOH would be 4 (use –log \[OH–\]) and pH would be 14–pOH = 14 – 4 = 10

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Question

What is the pH of a solution with \[OH-\] = 4 X 10-6

Answer

\[OH-\] = 4 X 10-6

pOH = 5.4 — use –log \[OH–\] to find pOH

pH = 14– pOH = 8.6

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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 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

Which of the following will produce the solution with the lowest pH?

Answer

NaOH is a base, so that won't produce an acidic solution. Of the remaining acids, HCl and HI are strong acids, and HF is weak. HI is at a higher molarity, so it will produce the most acidic solution.

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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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