200 mL of 0.10 M $NH X_{3}$ is mixed with 100 mL of 0.10 M HCl, and then 5 mmol of solid NaOH is dissolved in the mixture (no volume change). Given $p K_{b} = 4.74$ , $p K_{w} = 14$ , $lo g 2 = 0.301$ , the final pH is

Answer: D

NH3 = 20 mmol, HCl = 10 mmol gives NH3 = 10, NH4+ = 10 mmol. NaOH (5 mmol) converts NH4+ to NH3: NH3 = 15, NH4+ = 5 mmol. pOH =

4.74 + lo g (5/15) = 4.263

; pH = 9.74.

100 mL of a weak electrolyte solution is mixed with 100 mL of a strong electrolyte solution. Which of the following combinations give a buff... JEE Main level
A buffer is made with [CH3COOH] = 0.10 M and [CH3COO-] = 0.20 M (pK_a = 4.74, log 2 = 0.301). Select the correct statements. JEE Main level
100 mL of a buffer contains 0.20 M CH3COOH and 0.40 M CH3COONa (pK_a = 4.74). On adding 10 mL of 1.0 M HCl the pH becomes JEE Advanced level
A solution containing 0.10 M HA and 0.050 M NaA has a measured pH of 4.40 (log 2 = 0.301, 10^-4.40 = 3.98 x 10^-5). Assuming ideal behaviour... JEE Advanced level
What amount of solid NaOH (no volume change) must be added to 1.0 L of 0.20 M weak acid HA (pK_a = 4.74) to make a solution of pH 5.74? JEE Advanced level
100 mL of buffer P (0.20 M HA, 0.10 M NaA) is mixed with 300 mL of buffer Q (0.10 M HA, 0.30 M NaA). With pK_a(HA) = 4.76 and log 2 = 0.301,... JEE Advanced level