We are going to work the calculation of the pH of a buffer prepared by mixing a
ID: 1071420 • Letter: W
Question
We are going to work the calculation of the pH of a buffer prepared by mixing a strong base with a weak acid. We want you to see all the steps of this type of problem in one place. You can learn the pieces of this problem by just answering the questions in stages. Make sure you can do the moles, then check your answer. Then do the stoich check your answer. Think about all the pieces and where you are strong and where you need practice. This is a good pu it-all-together type of problem. Enjoy! Compute the ph of the solution after the reaction of 48mL of 0.73M HCN with L of 0.39 KOH. HCN (aq) KoHlaq) Compute the moles of reactants before reaction: Number Number Moles of acid: Moles of base: moles moles Moles of reactant and pH active product after reaction: Number Number Excess reactant Product moles moles Now convert moles to concentration units: Number Number Molar [Product] r Reactant] MolarExplanation / Answer
Moles = Molarity* Volume(L)
Moles :HCN = 0.73*48/1000=0.03504, KOH= 0.39*27/1000=0.01053
From the reaction HCN+ KOH-->HOH+ KCN
1 mole of HCN requires 1 mole of KOH. So all the KOH is cosumed since it is limiting.
So moles of KCN formed = 0.01053, moles of HCN remaining = 0.03504-0.01053=0.02451
Volume after mixing = 48+27= 75ml = 75/1000L=0.075L
Concentrations : KCN= 0.01053/0.075=0.1404M HCN= 0.02451/0.075=0.33M
HCN---à H++CN-
ICE tables
HCN H+ CN-
Initial 0.33 0 0
Change -x x x
Equilibrium 0.33-x x x
Ka= [H+][CN-]/[HCN] = x2/(0.33-x)= 3.5*10-4
Looking at Ka value, 0.33-x is approximated to 0.33
Hence x2=3.5*10-4*0.33, x=0.0107M
[H+] =0.0107, pH= 1.97
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