Question

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is the pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution?

Answer 1

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$.

(a)  Relation between $K_{a}$ and $pK_{a}$ is as follows.

                       $pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

                      $pK_{a} = -log (K_{a})$

                                    = $-log(4.5 \times 10^{-4})$

                                     = 3.347

Also, relation between pH and  $pK_{a}$ is as follows.

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

                     = $3.347+ log \frac{0.15}{0.12}$

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = $Molarity \times volume$

                                            = $11.6 M \times 0.001 L$

                                             = 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

                                           = 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

           $pK_{a} = -log (K_{a})$

                         = $-log(4.5 \times 10^{-4})$

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

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

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.