Calculate pH change when 0.01 mol CH 3 KCOO Na solution is added to 1 L of 0.01 M CH 3 COOH solution. Ka( CH 3 COOH )=1.8 × 10-5, pK a =4.74

Question

Calculate pH change when 0.01 mol CH 3 KCOO Na solution is added to 1 L of 0.01 M CH 3 COOH solution. Ka( CH 3 COOH )=1.8 × 10-5, pK a =4.74 (A) 3.37 (B) 1.37 (C) 4.74 (D) 8.01

Tardigrade Admin · Accepted Answer

pH of 0.01 MCH 3 COOH pH =(1/2)[ p K a - log C] =(1/2)(4.74- log 0.01) =(1/2)(4.74+2) =3.37 When 0.01 mol CH 3 COONa is added to it, it is now a buffer and [[ CH 3 COONa ]=0.01] M. Now from pH = p K a + log ([ CH 3 COO ]/[ CH 3 COOH ]) =4.74+ log (0.01/0.01)=4.74 ∴ Change in pH =4.74-3.37=1.37

Q. Calculate pH change when 0.01molCH3​KCOONa solution is added to 1L of 0.01MCH3​COOH solution. Ka​(CH3​COOH)=1.8×10−5,pKa​=4.74

3.37

1.37

4.74

8.01

pH of 0.01MCH3​COOH pH=21​[pKa​−logC] =21​(4.74−log0.01) =21​(4.74+2) =3.37 When 0.01molCH3​COONa is added to it, it is now a buffer and [[CH3​COONa]=0.01]M. Now from pH=pKa​+log[CH3​COOH][CH3​COO]​ =4.74+log0.010.01​=4.74 ∴ Change in pH=4.74−3.37=1.37

Q. Calculate $p H$ change when $0.01 m o l C H_{3} K COO N a$ solution is added to $1 L$ of $0.01 M C H_{3} COO H$ solution.
$K_{a} (C H_{3} COO H) = 1.8 \times 1 0^{- 5}, p K_{a} = 4.74$