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((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((1L)/(1000mL))((0.50mol)/(1L))`

= 0.0125 mol

moles of HCl =
`25mL((1L)/(1000mL))((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 =
`(0.0075mol)/(0.050L)`

= 0.15 M

concentration of ammonium ion in buffer =
`(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((0.15)/(0.10))`

pH = 9.25 + 0.18

pH = 9.43

So, the correct choice is e) 9.43 .