Question

A buffer is prepared by combining 25 ml of 0.5O MNH(aq) with 25 mL of 0.20 M HCI. What is the pH of the buffer? (KA (NH4) = 5.6 x 10 ) a. 8.86 b. 9.65 c. 8.00 d. 7.76 e. 9.43

Answer 1

Answer:  e) 9.43

Explanation: A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid and the pH of the buffer is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

There are 25 mL of 0.50 M $NH_3$ which is a weak base. 25 mL of 0.20 M HCl, a strong acid are added. Ammonia reacts with HCl to form its conjugate acid, ammonium ion. The net ionic equation will be:

$NH_3(aq)+H^+(aq)\rightarrow NH_4^+(aq)$

From above reaction, ammonia and HCl react in 1:1 mol ratio. Let's calculate the moles of each we have before the reaction. There is also, 1:1 mol ratio between HCl and ammonium ion.

moles of ammonia = $25mL(\frac{1L}{1000mL})(\frac{0.50mol}{1L})$

= 0.0125 mol

moles of HCl = $25mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})$

= 0.005 mol

Excess moles of ammonia = 0.0125 - 0.005 = 0.0075 mol

moles of ammonium ion formed = 0.005 mol

Total volume of the solution = 0.025 mL + 0.025 mL = 0.050 L

concentration of ammonia in buffer = $\frac{0.0075mol}{0.050L}$

= 0.15 M

concentration of ammonium ion in buffer = $\frac{0.005mol}{0.050L}$

= 0.10 M

pKa is calculated from given Ka as:

$pKa=-logKa$

$pKa=-log5.6*10^-^1^0$

pKa = 9.25

Plug in the values in Handerson equation:

$pH=9.25+log(\frac{0.15}{0.10})$

pH = 9.25 + 0.18

pH = 9.43

So, the correct choice is e) 9.43 .