*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **+ ], and [OH-] values for a 0.0525M solution of hydroioidic acid. HI is strong and ionizes about 100% so at equilibrium [H+] = 0.0525M so pH = 1.280 then pOH = 12.720 and [OH-] = 1.90 x 10-13 M 6. Calculate the pH for a 1.35 M solution of a weak acid HA with K a = 8.39 x 10-5 . This is a weak acid so equilibrium is reached. HA(aq) + H 2 O(l) ' A -(aq) + H 3 O + (aq) I 1.35 - 0 0 C -x - +x +x E (1.35-x) - (x) (x) K a = 8.39 x 10-5 = x 2 / (1.35-x) approx = x 2 / 1.35 8.39 x 10-5 (1.35) = x 2 take the square root of both sides x = 0.01064 which is also H+ so pH = -log x = 1.973 = pH check approximation 0.01064 / 1.35 x 100% = 0.788% which is less than 5% so valid...

View
Full
Document