PLEASE ANSWER EACH PART!!!!!!! HELP 1. We measure out 24.00 mL of a 1.50 M solut

Question

PLEASE ANSWER EACH PART!!!!!!! HELP

1. We measure out 24.00 mL of a 1.50 M solution of HCN for our titration with sodium hydroxide.

a. If you add 1.60 mL of a 0.70 M solution of sodium hydroxide, what is the pH of the acid solution at this stage of the titration? The Ka for HCN is 4.9 X 10-10.

b. How many mL of 0.70 M sodium hydroxide are needed to reach the endpoint?

c. What is the pH at the equivalence point of this titration? Remember that since the conjugate base, CN-, is hydrolyzing in solution, you will need a Kb value for this base.

d. If we add 61.43 mL of sodium hydroxide we will be well past the equivalence point of the titration. What will be the pH at this point?

1.60 mL of sodium hydroxide added to the weak acid
0.70M sodium hydroxide
in the buret as the titrant 24.0 mL of original 1.50M HCNsample

Explanation / Answer

Ka of HCN= 4.9*10-10

pKa= -log (4.9*10-10)= 9.31

Moles of HCN= molarity* Volume in Liters, 1000ml= 1L

Hence molarity of HCN= 1.5*24/1000=0.036 and moles of NaOH added = 0.7*1.6/1000= 0.00112

The reaction between HCN and NaOH is HCN+ NaOH ---->NaCN+ H2O

Theoretical molar ratio of HCN: NaOH= 1:1

Actual molar ratio of HCN: NaOH= 0.036:0.00112= 0.036/0.00112 : 0.00112/0.00112=32.14:1

So excess is HCN and all the NaOH react. Moles of NaCN formed =0.00112, moles of HCN remaining = 0.036-0.00112= 0.03488

Volume of solution after mixing = 24+1.6= 25.6 ml=25.6/1000 L

Concentration = moles/Liter

Concentrations: HCN= 0.03488/(25.6/1000), CN-( from NaCN)=0.00112/(25.6/1000)

Since pH= pKa+ log [CN-]/[HCN]=9.31+log (0.00112/0.03488) =7.82

2.

At equivalence point, moles of NaOH= moles of HCN= 0.036

Volume of 0.7M NaOH= 0.036/0.7= 0.0514L=51.4 ml

3.

At pH point, moles of HCN= moles of NaOH= moles of CN- formed =0.036

Volume of solution at equivalence point= 51.4+24= 75.4 ml= 75.4/1000L

Concentration of CN- = 0.036*1000/75.4=0.7M

CN- + H2O -----àHCN+OH-

Kb= 10-14/Ka= 10-14/(4.9*10-10)= 2.04*10-5= [HCN][OH-]/[CN-]

Let x= drop in concentration of CN- to reach equilibrium, at equilibrium

[CN-]=0.7-x, [HCN]=[OH-]=x

Hence x2/(0.7-x)= 2.04*10-5, when solved for x using excel, x= 0.00377

pOH= -log (OH-)= -log(0.00377)= 2.42, pH= 14-2.42= 11.58

4. moles of NaOH in 61.43ml= 0.7*61.43/1000 =0.043, moles of excess NaOH after reaction with HCN = moles of   NaOH- moles of HCN = 0.043-0.036=0.007

volume of solution after mixing = 61.43+24= 85.43ml=85.43/1000L= 0.08543L

concentration of NaOH= moles/Volume in L= 0.007/0.08543 M=0.082

NaOH being strong base ionizes completely and hence NaOH --------->Na++OH-

[OH-]= 0.082, pOH= 1.08, pH= 14-1.08= 12.92

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