Question

Part A) Calculate the pH of a weak acid HA. The solution concentration is 0.035 M in HA. Ka=4.5*10^-6

Part B) Calculate the pH of 0.00012 M KOH solution.

Answer 1

dissociation A for the weak acid HA the equation of can be written as HA A & A Initial 0.035 0 0 Change x x x Equilibrium (0.035-x) x x So, the dissociation constant is given by A ka UKO 72 x.x = 10:035-2)

Since HA is a very weak acid, x << 0.035 So, we can write, (01035-x) = 0.035 (neglectings ma 22. 0.035 This equation (i) becomes, ka= 0,035 x² = 0.035ka - x = 10.035 x ka 0.035 x 4.5 x 10-6 = 3.97 x 10-4 MI Now x is the concentration of Ht in equilibrium. So, Concentration of tt, [++] = 3.97 x 10-4M. Hence pH of the weak acid can be calculated pH = -lags [tt = log (3.97X10²4) it = 3:40 Hence, pH of the weak acid is pH = 3.40 (Aus) B) KOH is a strongs base. So, at equilibrium with completely dissociate. The equation is, kot kt + oH- Initial 0.00012 final o r 0.00012 0.00012

So, the concentration of ot, [OH-] = 0.00012 M The poth of kot can be calculated as, pat ar log [ort ] ar logo (0.00012) - 3.92 Now, we share, (since pkw = 14) pH + port = 14 So, pH 2 14-POH 14- 3.92 10:08 Hence, pH of 0.00012 M kot is 10.08 (Aus).