In a buffer solution concentration of (NH4)2SO4 and NH4OH are 0.4 M and 0.8 M respectively, pH of the solution is [ Ka(NH4+) = 10-8]

Question

In a buffer solution concentration of (NH4)2SO4 and NH4OH are 0.4 M and 0.8 M respectively, pH of the solution is [ Ka(NH4+) = 10-8] (A) 7.2 (B) 8 (C) 9.2 (D) 10

Tardigrade Admin · Accepted Answer

(Kb)NH4OH =(10-14/10-8)=10-6 [NH4+]=0.4×2=0.8 M [NH4OH]=0.8 M pOH=PKb+log( text[salt]/ text[base]) pOH=-log10-6+log(0.8/0.8)=6 PH=14-pOH=14-6=8

Q. In a buffer solution concentration of $(N H_{4})_{2} S O_{4}$ and $N H_{4} O H$ are $0.4 M$ and $0.8 M$ respectively, $p H$ of the solution is $[K_{a} (N H_{4}^{+}) = 1 0^{- 8}]$

7.2

8

9.2

10

$(K_{b})_{N H_{4} O H} = \frac{1 0 ^{- 14}}{1 0 ^{- 8}} = 1 0^{- 6}$
$[N H_{4}^{+}] = 0.4 \times 2 = 0.8 M$
$[N H_{4} O H] = 0.8 M$
$pO H = P K_{b} + l o g \frac{[ s a lt ]}{[base]}$
$pO H = - l o g 1 0^{- 6} + l o g \frac{0.8}{0.8} = 6$
$P H = 14 - pO H = 14 - 6 = 8$

Q. In a buffer solution concentration of (NH4​)2​SO4​ and NH4​OH are 0.4M and 0.8M respectively, pH of the solution is [Ka​(NH4+​)=10−8]

7.2

8

9.2

10

(Kb​)NH4​OH​=10−810−14​=10−6 [NH4+​]=0.4×2=0.8M [NH4​OH]=0.8M pOH=PKb​+log[base][salt]​ pOH=−log10−6+log0.80.8​=6 PH=14−pOH=14−6=8

Q. In a buffer solution concentration of $(N H_{4})_{2} S O_{4}$ and $N H_{4} O H$ are $0.4 M$ and $0.8 M$ respectively, $p H$ of the solution is $[K_{a} (N H_{4}^{+}) = 1 0^{- 8}]$

$(K_{b})_{N H_{4} O H} = \frac{1 0 ^{- 14}}{1 0 ^{- 8}} = 1 0^{- 6}$
$[N H_{4}^{+}] = 0.4 \times 2 = 0.8 M$
$[N H_{4} O H] = 0.8 M$
$pO H = P K_{b} + l o g \frac{[ s a lt ]}{[base]}$
$pO H = - l o g 1 0^{- 6} + l o g \frac{0.8}{0.8} = 6$
$P H = 14 - pO H = 14 - 6 = 8$