A) 3.4
B) 2.6
C) 1.5
D) 1.99
Correct Answer: D
Solution :
\[pOH=-\log {{K}_{b}}+\log \left[ \frac{salt}{base} \right]\] \[[14-pH]=-\log [1.8\times {{10}^{-5}}]+\left[ \frac{salt}{1.0} \right]\] \[5=4.7+\log \left[ \frac{salt}{1.0} \right]\] \[\log \left[ \frac{salt}{1.0} \right]=5-4.7=0.3\] \[[salt]=antilog\text{ }0.3\] \[[salt]=1.995\]You need to login to perform this action.
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