Determine -OH- of a 0-050 M solution of ammonia to which has been added sufficient NH-4Cl to the total -NH-4- equal to 0-100-K-b-NH-3-1-8 times 10-5-

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PKb-x27F9-Kb-1-8-xD7-10-x2212-5pKb-5-x2212-log-1-8-4-745pOH-pKb-logSaltbase-4-745-x2212-log0-1000-050-4-475-x2212-log2pOH-4-4436-OH-x2212-3-6-xD7-10-x2212-5M

Determine $[O H^{-}]$ of a 0.050 M solution of ammonia to which has been added sufficient $N H_{4} C l$ to make the total $[N H_{4}^{=}]$ equal to 0.100. $[K_{b (N H_{3})} = 1.8 \times 10^{- 5}]$

$p K_{b} ⟹ K_{b} = 1.8 \times 10^{- 5}$
$p K_{b} = 5 - l o g (1.8) = 4.745$
$p O H = p K_{b} + l o g \frac{S a l t}{b a s e} = 4.745 - l o g \frac{0.100}{0.050}$
$= 4.475 - l o g 2$
$p O H = 4.4436$
$[O H^{-}] = 3.6 \times 10^{- 5} M$

Determine [OH−] of a 0.050 M solution of ammonia to which has been added sufficient NH4Cl to make the total [NH=4] equal to 0.100.[Kb(NH3)=1.8×10−5]

pKb⟹Kb=1.8×10−5pKb=5−log(1.8)=4.745pOH=pKb+logSaltbase=4.745−log0.1000.050=4.475−log2pOH=4.4436[OH−]=3.6×10−5M

Determine $[O H^{-}]$ of a 0.050 M solution of ammonia to which has been added sufficient $N H_{4} C l$ to make the total $[N H_{4}^{=}]$ equal to 0.100. $[K_{b (N H_{3})} = 1.8 \times 10^{- 5}]$

$p K_{b} ⟹ K_{b} = 1.8 \times 10^{- 5}$
$p K_{b} = 5 - l o g (1.8) = 4.745$
$p O H = p K_{b} + l o g \frac{S a l t}{b a s e} = 4.745 - l o g \frac{0.100}{0.050}$
$= 4.475 - l o g 2$
$p O H = 4.4436$
$[O H^{-}] = 3.6 \times 10^{- 5} M$