Since $\displaystyle \text{HCl}$ is a strong acid, the concentration of $\displaystyle H_3O^+$ is equal to the concentration of the acid itself.

Thus, $\displaystyle [H_3O^+]=0.40M$.

Recall how to find the pH of a solution:

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

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

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

Remember to maintain the correct number of significant figures.

Start by assuming that there is $\displaystyle 1$ liter of the solution. From this, we can use the given density to find the mass of the solution.

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

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

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

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

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

Since we initially assumed that we had $\displaystyle 1$ liter of the solution, we now also know the concentration of $\displaystyle \text{HBr}$ in this solution.

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

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

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

Recall how to find the pH of a solution.

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

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

Start by assuming that there is $\displaystyle 1$ liter of the solution. From this, we can use the given density to find the mass of the solution.

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

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

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

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

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

Since we initially assumed that we had $\displaystyle 1$ liter of the solution, we now also know the concentration of $\displaystyle \text{HI}$ in this solution.

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

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

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

Recall how to find the pH of a solution.

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

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

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.