1 mL of 0.01 N HCl is added to 999 mL solution of 0.1 N Na2SO4. The pH of the resulting solution will

Question

1 mL of 0.01 N HCl is added to 999 mL solution of 0.1 N Na2SO4. The pH of the resulting solution will (A) 2 (B) 7 (C) 5 (D) 1

Tardigrade Admin · Accepted Answer

As Na 2 SO 4 solution is neutral, it simply dilutes the HCl solution from 1 mL to 1000 mL. Now [ H +] =(0.01/1000)=10-5 M ∴ pH =- log [ H +] =- log 10-5 =[-5 log 10] pH =5

Q. $1 m L$ of $0.01 N H Cl$ is added to $999 m L$ solution of $0.1 N N a_{2} S O_{4}$ . The $p H$ of the resulting solution will

2

7

5

1

$A s N a_{2} S O_{4}$ solution is neutral, it simply dilutes the $H Cl$ solution from $1 m L$ to $1000 m L$ .

Now $[H^{+}] = \frac{0.01}{1000} = 1 0^{- 5} M$

$∴ p H = - lo g [H^{+}]$

$= - lo g 1 0^{- 5}$

$= [- 5 lo g 10]$

$p H = 5$

Q. 1mL of 0.01NHCl is added to 999mL solution of 0.1NNa2​SO4​. The pH of the resulting solution will

2

7

5

1

AsNa2​SO4​ solution is neutral, it simply dilutes the HCl solution from 1mL to 1000mL. Now [H+]=10000.01​=10−5M ∴pH=−log[H+] =−log10−5 =[−5log10] pH=5

Q. $1 m L$ of $0.01 N H Cl$ is added to $999 m L$ solution of $0.1 N N a_{2} S O_{4}$ . The $p H$ of the resulting solution will

$A s N a_{2} S O_{4}$ solution is neutral, it simply dilutes the $H Cl$ solution from $1 m L$ to $1000 m L$ .

Now $[H^{+}] = \frac{0.01}{1000} = 1 0^{- 5} M$

$∴ p H = - lo g [H^{+}]$

$= - lo g 1 0^{- 5}$

$= [- 5 lo g 10]$

$p H = 5$