Question

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

Answer 1

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[H_3O^+])_(pH)=underbrace(-log_10K_a)_(pK_a)+log_10{[[A^-]]/([HA(aq)]}}`

And we fill in the values....

`[HA]=[H_3C-CO_2H]=0.10*mol*L^-1`

`[A^(-)]=[H_3C-CO_2^(-)]=0.10*mol*L^-1`

And `pH=4.76+log_10{(0.10*mol*L^-1)/(0.10*mol*L^-1)}`

`=4.76`...why? because `log_10{(0.10*mol*L^-1)/(0.10*mol*L^-1)}=log_(10}1=0`

And looking at the equation addition of acetic acid will marginally DECREASE `pH`, and addition of acetate will marginally INCREASE `pH`. Of course, if we dilute the solution, the ratio `log_10{[[H_3C-CO_2^(-)]]/([H_3C-CO_2H(aq)]]}` WILL remain unchanged, so `pH` should be constant.

Please go thru the calculations yourself to determine `pH` under the given scenarios. You find that `pH=pK_a` when `[HA]=[A^-]`...