The pH of the solution formed by mixing 20 m L of 0.05 M H 2 SO 4 with 5.0 m L of 0.45 M NaOH at 298 K is

Question

The pH of the solution formed by mixing 20 m L of 0.05 M H 2 SO 4 with 5.0 m L of 0.45 M NaOH at 298 K is (A) 6 (B) 2 (C) 12 (D) 7

Tardigrade Admin · Accepted Answer

0.05 M H 2 SO 4=0.05 × 2=0.1 N H 2 SO 4 0.45 M NaOH =0.45 × 1=0.45 N NaOH Meq. of H 2 SO 4=0.1 × 20=2 Meq. of NaOH =0.45 × 5=2.25 Meq, of NaOH left =2.25-2=0.25 Hence, resulting solution is alkaline in nature.

Q. The $p H$ of the solution formed by mixing $20 m L$ of $0.05 M H_{2} S O_{4}$ with $5.0 m L$ of $0.45 M N a O H$ at $298 K$ is

6

2

12

7

$0.05 M H_{2} S O_{4} = 0.05 \times 2 = 0.1 N H_{2} S O_{4}$
$0.45 M N a O H = 0.45 \times 1 = 0.45 N N a O H$
Meq. of $H_{2} S O_{4} = 0.1 \times 20 = 2$
Meq. of $N a O H = 0.45 \times 5 = 2.25$
Meq, of $N a O H$ left $= 2.25 - 2 = 0.25$
Hence, resulting solution is alkaline in nature.

Q. The pH of the solution formed by mixing 20mL of 0.05MH2​SO4​ with 5.0mL of 0.45MNaOH at 298K is

6

2

12

7

0.05MH2​SO4​=0.05×2=0.1NH2​SO4​ 0.45MNaOH=0.45×1=0.45NNaOH Meq. of H2​SO4​=0.1×20=2 Meq. of NaOH=0.45×5=2.25 Meq, of NaOH left =2.25−2=0.25 Hence, resulting solution is alkaline in nature.

Q. The $p H$ of the solution formed by mixing $20 m L$ of $0.05 M H_{2} S O_{4}$ with $5.0 m L$ of $0.45 M N a O H$ at $298 K$ is

$0.05 M H_{2} S O_{4} = 0.05 \times 2 = 0.1 N H_{2} S O_{4}$
$0.45 M N a O H = 0.45 \times 1 = 0.45 N N a O H$
Meq. of $H_{2} S O_{4} = 0.1 \times 20 = 2$
Meq. of $N a O H = 0.45 \times 5 = 2.25$
Meq, of $N a O H$ left $= 2.25 - 2 = 0.25$
Hence, resulting solution is alkaline in nature.