1.

pH of two solutions : I. 50 mL of 0.2 MHCl + 50 mL of 0.2 MHA (K_(a)=1.0xx10^(-5)) and II. 50 mL of 0.2 M HCl +50 mL of 0.2 M NaA will be respectively

Answer»

0.70 and 2.85
1 and 2.85
1 and 3
3 and 1

Solution :[HCl] in the mixture `=(50xx0.2)/(100)=0.1M`
In PRESENCE of STRONG acid HCl, the weak acid HA remains PRACTICALLY unionized. Hence
`[H^+]=[H^(+)]_(HCl)=0.1M,pH=1`
(II) `NaA+HCl""toNaCl+HA`
`[HA]=(50xx0.2)/(100)=0.1M`
No HCl is left. Hence `[H^(+)]`
`=sqrt(K_(a)C)=sqrt(1.0xx10^(-5)xx0.1)=1.0xx10^(-3),pH=3`


Discussion

No Comment Found

Related InterviewSolutions