CALCULATION OF PH AND CONCENTRATION OF [H+] OR [OH-]

A solution of 0.01M concentration of NH_(4)OH is 2.6% dissociated. Calculate [H^(+)], [OH^(-)],[NH_(4)^(+)], [NH_(4)OH] and pH of solution.

At 298 K, the pH of a lemon juice is 2.32. Calculate its [H_(3)O^(+)]" and "[OH^(-)].

The pH of a solution is 8.6. Calculate the OH^(-) ion concentration pH=8.6 pOH=5.4 -log[OH^(-)]=10^(-5.4) [OH^(-)]=10^(-6)xx10^(0.6)=10^(-6)xx anto log 0.6 [OH^(-)]=3.98xx10^(-6)

pH is the negative logarithm of H^(+) pH=log[H^(+)] HCllt H^(+)+Cl^(-) H_(2)OhArrH^(+)+OH^(-) K_(W)=[H^(+)][OH^(-)] K_(W) depend on the temperatue. With rise in temperature K_(W) increases. At 298 K, pH of pure water = 7 At 373 K, pH of the pure water is

pH is the negative logarithm of H^(+) pH=log[H^(+)] HCllt H^(+)+Cl^(-) H_(2)OhArrH^(+)+OH^(-) K_(W)=[H^(+)][OH^(-)] K_(W) depend on the temperatue. With rise in temperature K_(W) increases. At 298 K, pH of pure water = 7 At 373 K, pH of the pure water is