0.1 (M) HCl and 0.1 (M) H2SO4 each of volume 2ml are mixed and the volume is made up to 6 ml by adding 2 ml of 0.01 (N) NaCl solution. The pH of the resulting mixture is

Question

0.1 (M) HCl and 0.1 (M) H2SO4 each of volume 2ml are mixed and the volume is made up to 6 ml by adding 2 ml of 0.01 (N) NaCl solution. The pH of the resulting mixture is (A) 1.17 (B) 10 (C) 0.3 (D)  log 2 - log 3

Tardigrade Admin · Accepted Answer

Mili moles of H+=0.1 × 2+0.1 × 2 × 2=0.6 Total volume in m l=6 p H=- log 10[H+] =- log ((0.6/6))=- log 0.1=1

Q. $0.1 (M) H Cl$ and $0.1 (M) H_{2} S O_{4}$ each of volume $2 m l$ are mixed and the volume is made up to $6 m l$ by adding $2 m l$ of $0.01 (N) N a Cl$ solution. The $p H$ of the resulting mixture is

1.17

10

0.3

$lo g 2 - lo g 3$

Mili moles of $H^{+} = 0.1 \times 2 + 0.1 \times 2 \times 2 = 0.6$
Total volume in $m l = 6$
$p H = - lo g_{10} [H^{+}]$
$= - lo g (\frac{0.6}{6}) = - lo g 0.1 = 1$

Q. 0.1(M)HCl and 0.1(M)H2​SO4​ each of volume 2ml are mixed and the volume is made up to 6ml by adding 2ml of 0.01(N)NaCl solution. The pH of the resulting mixture is

1.17

10

0.3

log2−log3

Mili moles of H+=0.1×2+0.1×2×2=0.6 Total volume in ml=6 pH=−log10​[H+] =−log(60.6​)=−log0.1=1

Q. $0.1 (M) H Cl$ and $0.1 (M) H_{2} S O_{4}$ each of volume $2 m l$ are mixed and the volume is made up to $6 m l$ by adding $2 m l$ of $0.01 (N) N a Cl$ solution. The $p H$ of the resulting mixture is

$lo g 2 - lo g 3$

Mili moles of $H^{+} = 0.1 \times 2 + 0.1 \times 2 \times 2 = 0.6$
Total volume in $m l = 6$
$p H = - lo g_{10} [H^{+}]$
$= - lo g (\frac{0.6}{6}) = - lo g 0.1 = 1$