25 mol of formic acid `(HCO_(2)H)` is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The `K_(a)` of formic acid is

Four grams of NaOH solid are dissolved in just enough water to make I litre of solution. What is the [H^(+)] . of the solution ?

Four grams of NaOH solid are dissolved in just enough water to make I litre of solution. What is the [H^(+)] . of the solution ?

When 10^(-6) mole of a monobasic strong acid is dissolved in one litre of solvent, the pH of the solution is ..............

When 10^(-6) mole of a monobasic strong acid is dissolved in one litre of solvent, the pH of the solution is ..............

When 10^(-6) mole of a monobasic strong acid is dissolved in one litre of solvent, the pH of the solution is ..............

Freezing Point Depression: when 15.0 grams of ethyl alcohol (C_(2)H_(5)OH) is dissolved in 750 grams of formic acid, the freezing point of the solution is 7.20^(@)C . The freezing point of pure formic acid is 8.40^(@)C Evalulate K_(f) for formic acid. Strategy: First calculate the molality and the depression of the freezing point. Then solve the equation DeltaT_(f)=K_(f)m for K_(f) by substituting values for m and DeltaT_(f) .

A bufer solution was prepared by dissolving 0.05 mol formic acid & 0.06 mol sodium formate in enough water to make 1.0 L of solution. Ka for formic acid is 1.80 xx 10^(-4) If this solution were diluted to 10 times its volume, what would be the pH ?