Question

How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH to 4.95? 3.87 mL 2nd attempt 1st attempt C TRY AG VIEW SOLUTION 13 OF 22 QUESTIONS COMPLETED SAMSUNG

Answer 1

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

            Moles of acetic acid = 0.0100 M × 1.00 L

                                              = 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

                                           = 0.100 mol

When $HNO_{3}$ is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles $HNO_{3}$)

                                                = 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles $HNO_{3}$)

                                        = 0.100 mol - x

According to Henderson - Hasselbalch equation,

           pH = $pK_{a} + log\frac{[conjugate base]}{[weak acid]}$

             pH = $pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}$

           4.95 = 4.75 + $log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$

       $log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$  = 4.95 - 4.75

                                            = 0.20

$\frac{(0.100 mol - x)}{(0.0100 mol + x)}$  = antilog (0.20)

                                           = 1.6

Hence,    x = 0.032555 mol

Therefore, moles of $HNO_{3}$ = 0.032555 mol

volume of $HNO_{3}$ = $\frac{moles HNO_{3}}{molarity of HNO_{3}}$

                                = $\frac{0.032555 mol}{10.0 M}$

                                 = 0.0032555 L

or,                             = 3.25           (as 1 L = 1000 mL)

Thus, we can conclude that volume of $HNO_{3}$ added is 3.26 mL.