Consider a weak base ‘B’ of pK_b = 5.699. ‘x’ mL of 0.02 M HCl and ‘y’ mL of 0.02 M weak base ‘B’ are mixed to make 100 mL of a buffer of pH 9 at 25°C. The values of ‘x’ and ‘y’ respectively are : [Given : log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990]

Consider a weak base B of pKb = 5.699. x mL of 0.02 M HCl and y mL of 0.02 M weak base B are mixed to make 100 mL of a buffer of pH 9 at 25°C. The values of x and y respectively are

Q. Consider a weak base ‘B’ of pK_b = 5.699. ‘x’ mL of 0.02 M HCl and ‘y’ mL of 0.02 M weak base ‘B’ are mixed to make 100 mL of a buffer of pH 9 at 25°C. The values of ‘x’ and ‘y’ respectively are :

[Given : log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990]

(A)

x	y
42.7	57.3

(B)

x	y
14.3	85.7

(C)

x	y
85.7	14.3

(D)

x	y
11.1	88.9

Correct Answer: x = 14.3 mL, y = 85.7 mL

Explanation

For a basic buffer:

pOH = pK_b + log ([salt] / [base])

Given pH = 9, hence:

pOH = 14 − 9 = 5

So:

5 = 5.699 + log ([salt] / [base])

log ([salt] / [base]) = −0.699

[salt] / [base] = 1 / 5

Salt is formed by neutralisation of base with HCl.

Let moles of HCl = moles of salt = 0.02 × x/1000 Moles of base initially = 0.02 × y/1000

Remaining base after reaction:

= 0.02(y − x)/1000

Thus:

(0.02x) / (0.02(y − x)) = 1 / 5

x / (y − x) = 1 / 5

5x = y − x

y = 6x

Given total volume = 100 mL:

x + y = 100

x + 6x = 100

x = 14.3 mL , y = 85.7 mL

Explanation

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