30 cc of (M/3) HCl, 20 cc of (M/N) HNO3 and 40 cc of (M/4) NaOH solutions are mixed and the volume was made up to 1 dm3. The pH of the resulting solution is

Question

30 cc of (M/3) HCl, 20 cc of (M/N) HNO3 and 40 cc of (M/4) NaOH solutions are mixed and the volume was made up to 1 dm3. The pH of the resulting solution is (A) 1 (B) 3 (C) 8 (D) 2

Tardigrade Admin · Accepted Answer

Total milliequivalents of H + =30 × (1/3)+20 × (1/2)=20 Total milliequivalents of OH - =40 × (1/4)=10 Milli equivalence of H + left =20-10=10 ∴ [ H +]=(10/1000) g text ions / dm 3=10-2 ∴ pH =2

Q. $30 cc$ of $\frac{M}{3} H Cl, 20 cc$ of $\frac{M}{N} H N O_{3}$ and $40 cc$ of $\frac{M}{4}$ $N a O H$ solutions are mixed and the volume was made up to $1 d m^{3}$ . The $p H$ of the resulting solution is

1

3

8

2

Total milliequivalents of $H^{+}$

$= 30 \times \frac{1}{3} + 20 \times \frac{1}{2} = 20$

Total milliequivalents of $O H^{-}$

$= 40 \times \frac{1}{4} = 10$

Milli equivalence of $H^{+}$ left

$= 20 - 10 = 10$

$∴ [H^{+}] = \frac{10}{1000} g ions / d m^{3} = 1 0^{- 2}$

$∴ p H = 2$

Q. 30cc of 3M​HCl,20cc of NM​HNO3​ and 40cc of 4M​ NaOH solutions are mixed and the volume was made up to 1dm3. The pH of the resulting solution is

1

3

8

2

Total milliequivalents of H+ =30×31​+20×21​=20 Total milliequivalents of OH− =40×41​=10 Milli equivalence of H+ left =20−10=10 ∴[H+]=100010​g ions /dm3=10−2 ∴pH=2

Q. $30 cc$ of $\frac{M}{3} H Cl, 20 cc$ of $\frac{M}{N} H N O_{3}$ and $40 cc$ of $\frac{M}{4}$ $N a O H$ solutions are mixed and the volume was made up to $1 d m^{3}$ . The $p H$ of the resulting solution is

Total milliequivalents of $H^{+}$

$= 30 \times \frac{1}{3} + 20 \times \frac{1}{2} = 20$

Total milliequivalents of $O H^{-}$

$= 40 \times \frac{1}{4} = 10$

Milli equivalence of $H^{+}$ left

$= 20 - 10 = 10$

$∴ [H^{+}] = \frac{10}{1000} g ions / d m^{3} = 1 0^{- 2}$

$∴ p H = 2$