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Question

Chemistry Question on Equilibrium

Degree of dissociation of an acid HCl is 95% 0.192 g of the acid is present in 0.5 L of solution.The pH of the solution is

A

2

B

1

C

3

D

4

Answer

2

Explanation

Solution

No. of moles of HCl=0.19236.5=0.0053HCl=\frac{0.192}{36.5}=0.0053 Volume =0.5= 0.5 It Molarity of HCl=0.00530.5=1.06×10−2 MHCl =\frac{0.0053}{0.5}=1.06\times10^{-2}\,M [H+][H^{+}] in HclHcl solution =1.06×10−2×95100=1.06\times10^{-2}\times\frac{95}{100} =100.7×10−4=1×10−2=100.7\times10^{-4}=1\times10^{-2} Thus pH=−log[H]+pH=-log[H]^{+} =−log(1×10−2)=2=-log(1\times10^{-2})=2