Question

3) Hypochlorous acid (HOCI) is a weak monoprotic acid, with K. = 3.0 x 10-8 a) What is the value for pH for a 0.0264 M aqueous solution of hypochlorous acid? [12 points] b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypochlorous acid. What will be the pH for this new solution? [15 points)

Answer 1

Solution:- 3). a). We know that pH of weak acid is given by 1 1 pH = x p5, - x log [AH] Where, pKa =- log[Ka] =- log [3.0 x 10-S] =- [- 7.523] = 7.523, [AH] = 0.0264 M Putting these values in the above equation then we get 1 1 1 pH = 2 x 7.523 – 3 x log[0.0264] = x 7.523 – 3 x [-1.5784] pH = 3.761 +0.789 pH = 4.55 b). Balanced equation NaOH + HOCl → NaCl + HO 1 mole 1 mole 1 mole 1 mole First we need to calculate the mole of HOC1 Mole of HOCI = Molarity Volume = 0.0264 mole/L x 1 L = 0.0264 mole From balanced equation we kow that 1 mole of NaOH reacts with 1 mole of HOC1. So, 0.0100 mole of NaOH reacts with 0.0100 mole of HOC1. Hence the solution has excess mole of HOCI. Mole of HOCI in solution = Mole of HOCI - Mole of NaOH = 0.0264 -0.0100 = 0.0164 mole Volume of solution is 1L Molarity of solution = mole of HOCI Volume = 0.0164 mole/1 L = 0.0164 M We know that 1 1 x log [AH] pH =<xpKa = ax *7.523-x log[0.0164] = x 1 1 1 1 pH = x 7.523-5x [-1.7851] pH = 3.761 +0.892 pH = 4.65