Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

Question

Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is (A) 5.3 (B) 10.6 (C) 9.4 (D) 7.5

Tardigrade Admin · Accepted Answer

M(. H2 S O4 .)=(9.8/98 × 100)=10- 3m MN a O H=(4/40 × 100)=10- 3m Equivalents of resulting solution [. textO textH-].=(40 × 1 0- 3 - 10 × 1 0- 3 × 2/50)=(2/5)× 10- 3 textpOH=3. text39, textpH=10.6

Q. Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of $H_{2} S O_{4}$ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

5.3

10.6

9.4

7.5

$M_{(H_{2} S O_{4})} = \frac{9.8}{98 \times 100} = 1 0^{- 3} m$

$M_{N a O H} = \frac{4}{40 \times 100} = 1 0^{- 3} m$

Equivalents of resulting solution $[O H^{-}] = \frac{40 \times 1 0 ^{- 3} - 10 \times 1 0 ^{- 3} \times 2}{50} = \frac{2}{5} \times 1 0^{- 3}$

$pOH = 3. 39, pH = 10.6$

Q. Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H2​SO4​ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

5.3

10.6

9.4

7.5

M(H2​SO4​)​=98×1009.8​=10−3m MNaOH​=40×1004​=10−3m Equivalents of resulting solution [OH−]=5040×10−3−10×10−3×2​=52​×10−3 pOH=3.39,pH=10.6

Q. Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of $H_{2} S O_{4}$ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

$M_{(H_{2} S O_{4})} = \frac{9.8}{98 \times 100} = 1 0^{- 3} m$

$M_{N a O H} = \frac{4}{40 \times 100} = 1 0^{- 3} m$

Equivalents of resulting solution $[O H^{-}] = \frac{40 \times 1 0 ^{- 3} - 10 \times 1 0 ^{- 3} \times 2}{50} = \frac{2}{5} \times 1 0^{- 3}$

$pOH = 3. 39, pH = 10.6$