Question

A compound has a pka of 7.4. to 100 ml of a 1.0 m solution of this compound at ph 8.0 is added 30 ml of 1.0 m hydrochloric acid. the resulting solution is ph:

Answer 1

Explanation:

According to the Henderson-Hasselbalch equation for the solution before addition of HCl.

             pH = $pK_{a} + log(\frac{[A^{-}]}{[HA]})$

Now, putting the given values into the above formula as follows.

             8.0 = $7.4 + log(\frac{[A^{-}]}{[HA]})$

            $\frac{[A^{-}]}{[HA]}$ = 4.0   ............ (1)

Also, $0.1 L \times 1.0 mol/L$ = 0.1 moles is the total of the compound.

Therefore,  $[A^{-}] + [HA]$ = 0.1

                        $[A^{-}]$ = 0.1 - [HA] ............. (2)

Now, substituting the value of $[A^{-}]$ from equation (2) into equation (1) as follows.

                       $\frac{[A^{-}]}{[HA]}$ = 4.0

                      $\frac{0.1 - [HA]}{[HA]}$ = 4.0

                               0.1 - [HA] = 4.0[HA]

                             0.1 = 5.0[HA]

                           [HA] = 0.02 mol

Therefore, $[A^{-}]$ = 0.1 - [HA]

                                = 0.1 - 0.02

                                = 0.08 mol

Now, addition of strong acid reduces $A^{-}$ and increases AH by the same amount.

          (0.03 L) × (1 mol/L) = 0.03 moles HCl are added.

So,        $A^{-}$ = 0.08 - 0.03 = 0.05 moles

and,          AH = 0.02 + 0.03 = 0.05 moles

Therefore, after HCl addition, $\frac{[A^{-}]}{[HA]}$

                                = $\frac{0.05}{0.05}$

                                = 1.0

Now, re-substituting into the Henderson-Hasselbalch equation:

                           pH = 7.4 + log(1.0)

                                 = 7.4

Thus, we can conclude that pH of the resulting solution is 7.4.

Answer 2

 Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.