# A 75 mL solution that is 0.10 M in HC2H3O2 and 0.10 M in NaC2H3O2 has a pH of 4.74. Which of the following actions will change the pH of this solution?

## I. Adding 15 mL of 0.10 M HCl II. Adding 0.010 mol of NaC2H3O2 III. Diluting the solution from 75 mL to 125 mL The answer is I and II only. Could someone please explain why? Your help will be much appreciated! Thanks

Jan 7, 2018

You have prepared a buffer solution.

#### Explanation:

And that is a weak acid mixed with its conjugate base in appreciable concentrations. For which we use the buffer equation.....

${\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{p H} = {\underbrace{- {\log}_{10} {K}_{a}}}_{p {K}_{a}} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A \left(a q\right)\right]}\right\}$

And we fill in the values....

$\left[H A\right] = \left[{H}_{3} C - C {O}_{2} H\right] = 0.10 \cdot m o l \cdot {L}^{-} 1$

$\left[{A}^{-}\right] = \left[{H}_{3} C - C {O}_{2}^{-}\right] = 0.10 \cdot m o l \cdot {L}^{-} 1$

And $p H = 4.76 + {\log}_{10} \left\{\frac{0.10 \cdot m o l \cdot {L}^{-} 1}{0.10 \cdot m o l \cdot {L}^{-} 1}\right\}$

$= 4.76$...why? because ${\log}_{10} \left\{\frac{0.10 \cdot m o l \cdot {L}^{-} 1}{0.10 \cdot m o l \cdot {L}^{-} 1}\right\} = {\log}_{10} 1 = 0$

And looking at the equation addition of acetic acid will marginally DECREASE $p H$, and addition of acetate will marginally INCREASE $p H$. Of course, if we dilute the solution, the ratio ${\log}_{10} \left\{\frac{\left[{H}_{3} C - C {O}_{2}^{-}\right]}{\left[{H}_{3} C - C {O}_{2} H \left(a q\right)\right]}\right\}$ WILL remain unchanged, so $p H$ should be constant.

Please go thru the calculations yourself to determine $p H$ under the given scenarios. You find that $p H = p {K}_{a}$ when $\left[H A\right] = \left[{A}^{-}\right]$...