Your laboratory assistant is going to prepare a stock solution of sodium hydroxi
ID: 903863 • Letter: Y
Question
Your laboratory assistant is going to prepare a stock solution of sodium hydroxide (NaOH) for you. What mass of NaOH must the assistant add to a one liter volumetric so that the stock solution has a concentration of 0.35 M after dilution to the mark with water?
What is the pH of this stock solution?
You will now use this 0.35 M NaOH solution to carry out a series of dilutions.
Sample A will be produced by pipetting 1.00 mL of the stock solution into a 25 mL volumetric, and diluting to the volume with distilled water.
[OH-]=
[Na+]=
pH=
Sample B will be produced by pipetting 1.00 mL of sample A into a 25 mL volumetric, and diluting to volume with distilled water.
[OH-]=
[Na+]=
pH=
Sample C will be produced by pipetting 1.00 mL of sample B into a 25 mL volumetric, and diluting to volume with distilled water.
[OH-]=
[Na+]=
pH=
Sample D will be produced by pipetting 1.00 mL of sample C into a 25 mL volumetric, and diluting to volume with distilled water
[OH-]=
[Na+]=
pH=
Sample E will be produced by pipetting 1.00 mL of sample D into a 25 mL volumetric, and diluting to volume with distilled water
[OH-]=
[Na+]=
pH=
Sample F will be produced by pipetting 1.00 mL of sample E into a 25 mL volumetric, and diluting to volume with distilled water
[OH-]=
[Na+]=
pH=
Explanation / Answer
Concentration of stock=0.35M=0.35 moles/L=0.35 moles * molar mass of NaOH per L
=0.35 mol *40g/mol per L
=14g/L
pOH=-log[OH-]=-log(0.35)=0.45
pH=14-pOH=14-0.45=13.55
Sample A
Using equation ,M1 V1=M2 V2 ,
Where M1=molarity of solution being diluted
V1=volume of solution being diluted
M2=molarity of solution prepared
V2= volume of solution prepared
So we can write,M2=M1* V1/V2=0.35M *1.00ml/25 ml=0.014M
[OH-]=0.014M
[OH-]=[Na+]=[NaOH]
[Na+]=0.014M
pOH=-log[OH-]=-log(0.014)=1.85
pH=14-pOH=14-1.85=12.15
Sample B
M2=M1* V1/V2=0.014M *1.00ml/25 ml=0.00056M
[OH-]=0.00056M
[Na+]=0.00056M
pOH=-log[OH-]=-log(0.00056)=3.25
pH=14-pOH=14-3.25=10.75
Sample C
M2=M1* V1/V2=0.00056M *1.00ml/25 ml=2.24*10^-5 M
[OH-]=2.24*10^-5 M
[Na+]=2.24*10^-5 M
pOH=-log[OH-]=-log(2.24*10^-5)=4.65
pH=14-pOH=14-4.65=9.35
Sample D
M2=M1* V1/V2=(2.24*10^-5)M *1.00ml/25 ml=9.0*10^-7 M
[OH-]=9.0*10^-7 M
[Na+]=9/0 *10^-7 M
pOH=-log[OH-]=-log(9.0*10^-7)=6.05
pH=14-pOH=14-6.05=7.95
Sample E
M2=M1* V1/V2=(9.0*10^-7)M *1.00ml/25 ml=3.6 *10^-8 M
[OH-]=3.6 *10^-8 M
[Na+]=3.6 *10^-8 M
pOH=-log[OH-]=-log(3.6 *10^-8)=7.45
pH=14-pOH=14-7.45=6.55
Sample F
M2=M1* V1/V2=( 3.6 *10^-8)M *1.00ml/25 ml=1.4 *10^-9 M
[OH-]=1.4 *10^-9 M
[Na+]=1.4 *10^-9 M
pOH=-log[OH-]=-log(1.4*10^-9)=8.85
pH=14-pOH=14-8.85=5.15
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