Chemistry 2 50.0 mL of a solution of 0.20M HCOOH (K_a = 1.8 times 10^4) is titra

Question

Chemistry 2 50.0 mL of a solution of 0.20M HCOOH (K_a = 1.8 times 10^4) is titrated against 0.30M solution of KOH. What is the pH after the addition of 60.0 mL of KOM? 0.133 0.875 13.125 1.14 12.86 Show Work: Three questions worth 16 points total. Clearly write the equation and show the steps for all questions Determine the pH of a 0.43 M C_5H_5NHCl (pyridinium chloride) solution. The K_b of C_5H_5N (pyridine, a weak base) is 1.7 times 10^-9. The relevant equation is: C_5H_5NH^+(aq) + H_2O(l) C_5H_5N(aq) + H_3O^+(aq) Determine the molar solubility of Ca_3(PO_4)_2 in a solution containing 0.200 M K_3PO_4. The K_sp for Ca_3(PO_4)_2 is 1.2 times 10^-26.

Explanation / Answer

Q22.

V = 50 mL of acid

M = 0.5 M of acid

Ka = 1.8*10^-4 of acid

M = 0.3 M of base

V = 60 mL of base

so..

pH at this point:

mmol of acid = MV = 50*0.20 = 10 mmol of acid

mmol of base = MV = 60*0.3 = 18 mmol of base

note thta we are over the tequivalnece point, so expect a very high pH due to OH- presence

mmol of base left = 18-10 = 8 mmol of base left

Vtotal = 50+60 = 110 mL

[OH-] = mmol / mL = 8/110 = 0.07272 M

pOH = -log(OH) = -log(0.07272) = 1.1383

pH = 14-pOH = 14-1.1383 = 12.8617

beswt answer is E

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