Pure water is neutral in nature, meaning [ H + ] = [ OH − ] . When this condition is disturbed by changing the concentration of H + or OH − , the neutral solution becomes acidic {[ H + ] [ OH − ]} o

Question

Pure water is neutral in nature, meaning $[H^{+}] = [OH^{-}]$ . When this condition is disturbed by changing the concentration of $H^{+}$ or $OH^{-}$ , the neutral solution becomes acidic ${[H^{+}] > [OH^{-}]}$ or basic ${[H^{+}] < [OH^{-}]}$ . This change occurs during salt hydrolysis. Here,

$p K_{C H_{3} COO H} = 4.74$ .

When 50 mL of 0.1 M NaOH is added to 50 mL of 0.1 M CH₃COOH solution, the pH will be:

TestHub · Accepted Answer

NaOH + C H 3 ​ COOH ⇌ C H 3 ​ COONa + H 2 ​ O V 1 ​ = 50 ml V 2 ​ = 50 ml M 1 ​ = 0.1 M M 2 ​ = 0.1 M Since the millimoles of acid equal the millimoles of base, the pH will be calculated by the hydrolysis of the salt (formed from a weak acid and a strong base). ∴ ​ p H p H pH ​ = 7 + 2 1 ​ p K a ​ + 2 1 ​ lo g C = 7 + 2 1 ​ × 4.7447 + 2 1 ​ lo g 5 × 1 0 − 2 = 7 + 2.37 − 0.65 ​ [ C = 100 5 ​ ] [ lo g 5 = 0.6990 ] ​

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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