I\'ve tried 0.48, 0.5, 0.484, 13.5, 11.4, 11.36, 8.53, if you think the answer i

Question

I've tried 0.48, 0.5, 0.484, 13.5, 11.4, 11.36, 8.53, if you think the answer is one of these pleases don't respond to this question because it is not!!!!!

The reaction of the weak acid HCN with the strong base KOH is: HCN (aq) KoHlaq) HoH(I) KCN (aq) To compute the pH of the resulting solution if 62mL of 0.55M HCN is mixed with 2.0 x 101mL of 0.36 KOH we need to start with the stoichiometry. Let's do just the stoich in steps:. Number Number How many moles of acid? 0341 How many moles of base? 0072 KOH What is the limiting reactant? Number .0269 How many moles of the excess reactant after reaction? Number 0.328 What is the concentration of the excess reactant after reaction? Number What is the concentration of the pH active product after reaction? 8.53 These are the initial concentration in an ICE Table! Incorrect.

Explanation / Answer

This will form a buffer since we have

mmol of acid = MV = 62*0.55 = 34.1 mmol of acid

mmol of base = MV = 0.36*20 = 7.2 mmol of base

so

mmol of acid left after reaction = 34.1-7.2 =26.9 mmol of HCN left

mmol of conjguate base, CN- formed = 0 + 7.2 = 7.2 mmol ofCN-

this is a buffer

so

apply buffer equation

pH = pKa+ log(CN-/HCN)

pKa = -log(Ka) = -log(6.17*10^-10) = 9.209714

pH = 9.21 + log(7.2/34.1)

pH = 8.5342

So for concentration of H+

[H+] = 10^-pH = 10^-8.5342 = 0.000000002922806 = 2.9*10^-9

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