Pure water is netural in nature `[H^(+)]=[OH^(-)].` When this condition is disturbed by changing the concentration of `H^(+)or OH^(-),` the natural solution changes to acidic `{[H^(+)]gt[OH^(-)]}` or basic `{[H^(+)]lt[OH^(-)]}.` This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation
(i) Salt of weak acid and strong base
`pH =1/2[pK_(w)+pK_(a)+logC]`
(ii) Salt of weak base and strong acid
`pH=1/2[pK_(w)-pK_(b)-logC]`
(iii) For salt of weak base and strong acid
`pH=1/2[pK_(w)+pK_(a)-pK_(b)]`
The pH of buffer can be calculated using t he following formula
`pH=pK_(a)+log""(["Salt"])/(["Acid"])`
`pOH=pK_(b)=log""(["Salt"])/(["Base"])`
Answer t he following questions when
`pK_(a)=4.7447`
`pK_(b)=4.7447` ltb rgt `pK_(w)=14`
1 mole of `CH_(3)COOH` is dessolved in water to from 1 litre aqueous solution. The pH of resulting solution will be

Options

A. `9.2253`
B. `2.3723`
C. 14
D. 7

Answer

B