Question

U Experiment 17B: Acid-Base Titration Report Data Table Trial 1 Trial 2 T Trial 3 Volume of Acetic Acid Final buret reading 42.35 27.67 Initial buret reading 30.00 38.32 5.81 32.51 30.00 44.40 3.95 40.45 Volume of NaOH 27.67 Exact Molarity of NaOH (From 17A or see label) 1252 Calculations: 1. Calculate the molarity of the acetic acid for each trial and the average molarity Trial 1: Trial 2: Trial 3: Average Molarity of Acetic Acid = 2. Use the average molarity of the acetic acid from Q1 to calculate the % Acetic Acid. (Density of acetic acid is 1.01 17B-3

Answer 1

1. Molarity of Acetic acid(M₁) of each trial is given below

Molarity Of NaOH (M₂)= 0.252M

Volume of NaOH in each Trial is= V₂

Volume of Acetic Acid in each Trial is= V₁

Now use M₁V₁ =M₂V₂ , So M₁=M₂V₂/V₁

For trial I

Molarity of Acetic acid(M₁) = 0.252M*27.67mL/42.35mL = 0.165M

For trial II

Molarity of Acetic acid(M₁) = 0.252M*32.51mL/30.00mL = 0.273M

For trial III

Molarity of Acetic acid(M₁) = 0.252M*40.45mL/30.00mL = 0.340M

AverageMolarity of Acetic acid = (0.165M+0.273M+0.340M)/3 = 0.256M

AverageMolarity of Acetic acid = 0.256M

2.Percentage of Acetic acid

Density = 1.01g/mL
Average Molarity of Acetic acid = 0.256M

We know that for saturated Acetic acid 99.5% molarity is 17.4M , which having 1.01g/mL density

Note: You also take this data from any standard reference book.

So % acetic acid is = 0.256M*100/17.4M = 1.47%

3. Actual molarity of acetic acid is not given

When you obtain ,

Then use this formula to calculated % error of acetic acid molarity

% error of acetic acid molarity =(Actual molarity - Experimental molarity)*100/Actual molarity

4. If calculation is not in 5 % error of acetic acid molarity

Then your NaOH solution Molarity may not correct due to preparation of this is long time that may react with CO₂ in the atmosphere

which act as acid and react lit bit to NaOH to give carbonic acid. So change the concentration of NaOH , So prepare fresh solution of NaOH

POST LAB QUESTIONS

1.

Molarity of HBr (M₁) =?

Molarity Of NaOH (M₂)= 0.150M

Volume of NaOH(V₂) is = 18.8mL

Volume of HBr(V₁) = 25.00mL

Now use M₁V₁ =M₂V₂ , So M₁=M₂V₂/V₁

Molarity of HBr (M₁) = 0.150M*18.8mL/25.00mL = 0.1128M

So, Molarity of HBr (M₁) = 0.1128M

2.

Molarity of H₂SO₄ (M₁) =?

Molarity Of NaOH (M₂)= 0.100M

Volume of NaOH(V₂) is = 45.65mL

Volume of H₂SO₄(V₁) = 20.00mL

Now use M₁V₁ =M₂V₂ , So M₁=M₂V₂/V₁

Molarity of HBr (M₁) = 0.100M*45.65mL/20.00mL = 0.22825M

So, Molarity of H₂SO₄ (M₁) = 0.22825M