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