By adding 20 ml 0.1 N HCl to 20 ml 0.001 N KOH, the pH of the obtained solution will be

Question

By adding 20 ml 0.1 N HCl to 20 ml 0.001 N KOH, the pH of the obtained solution will be (A) 7 (B) zero (C) 1.3 (D) 4.2

Tardigrade Admin · Accepted Answer

Milliequivalents of acid =N1V1=20 × 0.1 N=2 Milliequivalents of base =N2V2=20× 0.001 N =0.02 Remaining acid = 2 - 0.02 = 1.8 [H+]=(1.8/V1+V2) =(1.8/40) =0.0045 M PH=-log[H+]=1.35

Q. By adding $20 m l 0.1 N H Cl$ to $20 m l 0.001 N K O H$ , the $p H$ of the obtained solution will be

$7$

zero

$1.3$

$4.2$

Milliequivalents of acid $= N_{1} V_{1} = 20 \times 0.1 N = 2$
Milliequivalents of base $= N_{2} V_{2} = 20 \times 0.001 N$
$= 0.02$
Remaining acid $= 2 - 0.02 = 1.8$
$[H^{+}] = \frac{1.8}{V _{1} + V _{2}}$
$= \frac{1.8}{40}$
$= 0.0045 M$
$P H = - l o g [H^{+}] = 1.35$

Q. By adding 20ml0.1NHCl to 20ml0.001NKOH, the pH of the obtained solution will be

7

zero

1.3

4.2

Milliequivalents of acid =N1​V1​=20×0.1N=2 Milliequivalents of base =N2​V2​=20×0.001N =0.02 Remaining acid =2−0.02=1.8 [H+]=V1​+V2​1.8​ =401.8​ =0.0045M PH=−log[H+]=1.35

Q. By adding $20 m l 0.1 N H Cl$ to $20 m l 0.001 N K O H$ , the $p H$ of the obtained solution will be

$7$

$1.3$

$4.2$

Milliequivalents of acid $= N_{1} V_{1} = 20 \times 0.1 N = 2$
Milliequivalents of base $= N_{2} V_{2} = 20 \times 0.001 N$
$= 0.02$
Remaining acid $= 2 - 0.02 = 1.8$
$[H^{+}] = \frac{1.8}{V _{1} + V _{2}}$
$= \frac{1.8}{40}$
$= 0.0045 M$
$P H = - l o g [H^{+}] = 1.35$