10. Determine the pH of 150 ml of a solution of a household bleach that contains

Question

10. Determine the pH of 150 ml of a solution of a household bleach that contains 4% (w/V) NaClO sodium hypochlorite if the pKa of hypochlorous acid is 7.52 11. A 0.85 M solution of sodium fluoride (NaF) has a pH of 8.54. Determine the hydrolysis constant for the NaF and the Ka for HF from this data. 12. what is the new pH of a solution if 10.0 ml of 0.10 M HC? is added to 150 ml of a buffer system that is 0.10 M Benzoic acid and 0.10 M sodium benzoate. (The Ka of benzoic acid is 6.3 x 10) 13. Phenobarbital (HC2HI,N203), a barbiturate drug is a nonselective central nervous system depressant, used primarily as a sedative and also as an anticonvulsant. It is generally delivered as the sodium salt, which is more soluble. When a tablet containing 50 mg of the salt, (NaC2H,N,O3), is dissolved in 400 ml of solution, the ph is determined to be 9.31. What is the Ka of the acid form of the drug (HC12H1,N203)?

Explanation / Answer

Volume of solution= 150ml, concentration of NaClO= 4% (W/V)

Mass of NaClO= 150ml* 4/100 = 6 gm

Moles of NaCl= mass/molar mass =6/74.5=0.081

Concentration of NaClO= moles/Volume= 0.081/(150/1000) =0.54M

NaClO is basic salt from reaction of weak acid HOCl and strong base, NaOH.

The hydrolysis of NaClO is ClO-+ H2O ----->HOCl+ OH-

Kb= [HOCl][OH-]/[ClO-], let x= drop in concentration of NaClO( from ClO-) to reach equilibrium. At Equilibrium, [OH-]=[ClO-]=x, and [ClO-]=0.54-x, Kb= 10-14/ Ka, Ka= 10(-7.2) =6.31*10-8

Kb= x2/(0.54-x)= 6.31*10-8, when solved for x, x=[OH-]= 1.846*10-4, pOH= -logx = 3.73

pH= 14-pOH= 14-3.73=10.27

2.

NaF is basic salt formed from reaction of weak acid HF with strong base, NaOH. Given pH= 8.54, poH= 14-pH= 14-8.54= 5.46, [OH-]= 10(-5.46)= 3.46*10-6,

NaF undergoes hydrolysis as NaF+ H2O ----->HF+ OH-, given [OH-]= [HF]=3.46*10-6

Kb= [HF][OH-]/[NaF]

Let x= drop in concentration of NaF to reach equilibrium.

At Equilibrium, [NaF]=0.45-x , x= 3.46*10-6, hence 0.45-x can be approximated to 0.45

Kb= 3.46*10-6*3.46*10-6/0.45 =2.671*10-11, Ka= 10-14/Kb= 10-14/(2.671*10-11)= 0.000374

3.

Moles = Molarity* volume in liters, volume of buffer= 150ml= 150/1000L=0.15L

Moles in the buffer : Benzoic acid (C6H5COOH) = 0.1*0.15=0.015, sodium Benzonate (C6H5COONa)= 0.1*0.15=0.015

Let HA is Benzoic acid and NaA is sodium Benzoate.

Moles of HCl added= molarity* volume inliters=0.1*10/1000=0.001

The addition of HCl leads to the reaction C6H5COONa + HCl ------àC6H5COOH. This gives more moles of Benzoic acid present in the solution. Theoretical ratio of C6H5COOH :HCl =1:1

Actual ratio of Benzoic acid and HCl supplied= 0.015:0.001, So HCl is limiting and all the HCl reacts. Moles of total Benzoic acid (HA)=0.001+0.015=0.016, moles of sodium benzonate (NaA)=0.015-0.001 =0.014

Volume of solution after mixing =150+10= 160ml= 160/1000L=0.16L

Concentrations : HA= 0.016/0.16=0.1, NaA= 0.014/0.16=0.0875

Now From Henderson-Hasselbalch equation

pH= pKa+ log [Base]/[Acid], given Ka= 6.3*10-5, pKa=-log Ka=4.2

hence pH= 4.2+log (0.0875/0.1)= 4.14

4.

Molar mass of Phenobarbital (C12H18N2O3)= 12*12+18*1+14*2+3*16=238 g/mole

Molar mass of the salt ( NaC12H17N2O3)= 23+12*12+17*1+14*2+3*16= 260 g/mole

Moles of salt= mass/molar mass = 50*10-3/260 =0.00019

This is dissolved in water of 400ml= 400/1000L=0.4L

Concentration of salt= 0.00019/0.4=0.00048M

Let the salt be represented as NaA and acid as HA. The NaA undergoes hydrolysis as

NaA+ H2O ----->HA+ OH-

Given pH= 9.31, pOH= 14-9.31=4.69, [OH-]= 10(-4.69)=2.04*10-5

Kb =[HA][OH-]/[NaA]

At Equilibriuml, [HA] =[OH-]= 2.04*10-5, NaA= 0.00048-2.04*10-5 =0.00046

Kb= 2.04*10-5*2.04*10-5/0.00046 =9.07*10-7, Ka=10-14/Kb= 10-14/(9.07*10-7)=1.1*10-8

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