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When `NH_(3)(0.1 M) 50 ml` mix with `HCl (0.1 M) 10 ml` then what is `pH` of resultant solution (`pK_(b)=4.75)`A. `9.25`B. `10`C. `9.85`D. `4.15` |
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Answer» Correct Answer - C `underset(rem.)underset()underset(Initial)(NH_(3))+underset(4mmol)underset(5mmol)underset(50xx0.1)(HCl) rarr underset(0)underset(1mmol)underset(10xx0.1)(NH_(4)Cl)` `pOH=pK_(b)+log``("salt")/(base)` `=4.75+log``(1)/(4)=4.15` `pH=14-pOH=14-4.15=9.85` |
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