Question

chans the 7. Determine the pH of the following solutions a. 0.040 M HCL ne b. 0.040 M KOH C. 0.0087 M HClO2 d. 0.010 M CH3NH2 (methylamine) e. 0.018 M NaF 0.10

Answer 1

a. He is a strong acid. Thus, the concentration of H+ in 0.040 M Hul is- [H+] = 0.040 M pH = -log [H+] = - Log (0.040) = 1.398 No ř 1.4 two significant figures i. pH = (1.4 Ans b. KoH is a strong base. The concentration of hydroxyl in in 0.040 M KOH is - [ot] = 0.040 pH = -log 0.040 = -1.398 pH = 14-pon = 14-1.398 = 12.6 i. pH = 12.6 | Aus Again,

& Hd Og is a weak acid. pka of HelO2. pka = 1.96 concentration, c = 0.0087 M Thus pH ton weak acid can be given as - pH = 2 pka - 1/2 loge - (1.96) - { log 0.0087 = 0.98 + 1.03 = 2.01 z 2 u. (pH = 2 An d. CH₂NH2 is a weak base. pko = 3.28 ; c= 0.010 Thus, port of weak base can be given pot = 2 pkb - 11 loge = 3-28 - Ilog 0.01

= 1.64 + 1 = 2.64 ...pH = 14- POH = 14 -2.64 = 11.4 pH = 11.47 Ang ļ. NaF is a salt produced from the weak acid HF and strong base NaOH For this type of salt pH can be given as - pH = 7+ 1 pka + 2 loge Now, pka of HF = 3.45 Concentration, c=0.018. pH = 7+ 1 / 2 x 3.45 + 1/2 log 0.018 = 7+ 1.725 - 0.872 = 7.853 7 7.8 pH = 7.8 Ane.