Calculate pH of 0.002 N NH 4 OH having 2 % dissociation

Question

Calculate pH of 0.002 N NH 4 OH having 2 % dissociation (A) 7.6 (B) 8.6 (C) 9.6 (D) 10.6

Tardigrade Admin · Accepted Answer

NH 4 OH is a weak base and partially dissociated <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/cf4f26d24842d36571ae4d9699c0e128-.png" /> ∴ [O H-] =C α=2 × 10-3 × (2/100)=4 × 10-5 M pOH =- log [ OH -] =- log 4 × 10-5=4.4 pH =14-4.4=9.6

Q. Calculate pH of $0.002 N N H_{4} O H$ having $2%$ dissociation

7.6

8.6

9.6

10.6

$N H_{4} O H$ is a weak base and partially dissociated

$∴ [O H^{-}] = C α = 2 \times 1 0^{- 3} \times \frac{2}{100} = 4 \times 1 0^{- 5} M$

$pO H = - lo g [O H^{-}]$

$= - lo g 4 \times 1 0^{- 5} = 4.4$

$p H = 14 - 4.4 = 9.6$

Q. Calculate pH of 0.002NNH4​OH having 2% dissociation

7.6

8.6

9.6

10.6

NH4​OH is a weak base and partially dissociated ∴[OH−]=Cα=2×10−3×1002​=4×10−5M pOH=−log[OH−] =−log4×10−5=4.4 pH=14−4.4=9.6

Q. Calculate pH of $0.002 N N H_{4} O H$ having $2%$ dissociation

$N H_{4} O H$ is a weak base and partially dissociated

$∴ [O H^{-}] = C α = 2 \times 1 0^{- 3} \times \frac{2}{100} = 4 \times 1 0^{- 5} M$

$pO H = - lo g [O H^{-}]$

$= - lo g 4 \times 1 0^{- 5} = 4.4$

$p H = 14 - 4.4 = 9.6$