A solution obtained on mixing 0.1 N , 50 mL HCl and 0.1 N , 50 mL NaOH. The value of pH of the solution will be

Question

A solution obtained on mixing 0.1 N , 50 mL HCl and 0.1 N , 50 mL NaOH. The value of pH of the solution will be (A) 4 (B) 6 (C) 8 (D) 7

Tardigrade Admin · Accepted Answer

HCl+NaOH xrightarrow[]NaCl+H2O Equivalent number of HCl=(0.1× 50/1000)=0.005 Equivalent number of NaOH =(0.1× 50/1000)=0.005 Equivalent of HCl= Equivalent of NaOH 0.005≡ 0.005 Hence, solution is neutral ie., pH=7

Q. A solution obtained on mixing $0.1 N, 50 m L$ $H Cl$ and $0.1 N, 50 m L N a O H$ . The value of $p H$ of the solution will be

4

6

8

7

$H Cl + N a O H N a Cl + H_{2} O$ Equivalent number of $H Cl = \frac{0.1 \times 50}{1000} = 0.005$ Equivalent number of $N a O H$ $= \frac{0.1 \times 50}{1000} = 0.005$ Equivalent of $H Cl =$ Equivalent of $N a O H$ $0.005 \equiv 0.005$ Hence, solution is neutral ie., $p H = 7$

Q. A solution obtained on mixing 0.1N,50mL HCl and 0.1N,50mLNaOH. The value of pH of the solution will be

4

6

8

7

HCl+NaOH​NaCl+H2​O Equivalent number of HCl=10000.1×50​=0.005 Equivalent number of NaOH =10000.1×50​=0.005 Equivalent of HCl= Equivalent of NaOH 0.005≡0.005 Hence, solution is neutral ie., pH=7

Q. A solution obtained on mixing $0.1 N, 50 m L$ $H Cl$ and $0.1 N, 50 m L N a O H$ . The value of $p H$ of the solution will be

$H Cl + N a O H N a Cl + H_{2} O$ Equivalent number of $H Cl = \frac{0.1 \times 50}{1000} = 0.005$ Equivalent number of $N a O H$ $= \frac{0.1 \times 50}{1000} = 0.005$ Equivalent of $H Cl =$ Equivalent of $N a O H$ $0.005 \equiv 0.005$ Hence, solution is neutral ie., $p H = 7$