Question

What is the pH of a solution made by mixing 100.0 mL of 0.10 M $HN{O}_3$, 50.0 mL 0f 0.20 HCl and 100.0 mL of water? Assume that the volumes are additive.

• A. $2.7$
• B. $1.10$
• C. $3.50$
• D. $1.5$

Answer 1

The correct option is B $1.10$

$V_{total}=100.0+50+100=250$ mL

Now, we are dealing with two strong acids that ionize completely in aqueous solution. Both nitric acid and hydrochloric acid produce hydronium cations in 1:1 mole ratios.

$[H_3{O}^{+}{]}_{comingfromHN{O}_3}=[HN{O}_3]$

$[H_3{O}^{+}{]}_{comingfromHCl}=[HCl]$

We know, molarity is defined as the number of moles of solute present in ${10}^3$ mL of solution. For the nitric acid solution, we have,

$100.0mL=\frac{{10}^{3}mL}{10}$

$n_{H_3{O}^{+}}=\frac{0.10moles}{10}=0.010moles{H}_3{O}^{+}$

For hydrochloric acid,

$50.0mL=\frac{{10}^{3}mL}{20}$

$n_{H_3{O}^{+}}=\frac{0.20moles}{20}=0.010moles{H}_3{O}^{+}$

The total number of moles of hydronium cations delivered by the two acids in the resulting solution will be :

$n_{H_3{O}^{+}}=0.010+0.010=0.020moles$

The concentration of the hydronium cations in the resulting solution will be :

$\left[H_3{O}^{+}\right]=\frac{0.020moles}{250.0\times {10}^{-3}L}=0.080M$

As you know,

$pH=-\log \left(H_3{O}^{+}\right)=-\log \left(0.080\right)=1.10$