Answer to Question #345627 in General Chemistry for kris

Question

tritrations were performed on a 300.00mL samples of water taken from Lake Ontario. 4 trials were done. 54.5 mL/35.8 mL/ 53.8 mL/54.1 mL of 0.020 mol/L KOH was used to neutralize the sulphuric acid (H2SO4) in the lake water sample. Calculate the concentration of the phosphoric acid in the lake.

Answer 1

H₂SO₄ + 2KOH = K₂SO₄ + 2H₂O

1. We find the amount of substance (mol) of KOH used:

n = C_M * V in the formula,

n — amount of substance (mol);

C_M — concentration of molarity (mol/l) = =0,02 mol/l;

V — volume of solution (l) .

2. According to the above reaction, 1 mole of H₂SO₄ reacts with 2 moles of KOH. therefore we divide the amount of KOH (mol) by 2. resulting in an amount (mole) of H₂SO₄

3. Divide the result by the volume of the water sample being tested (0.3 l). resulting in a molar concentration of H₂SO₄

We calculate the above steps for each trials

a)

1. n = 0,02 * 0,0545 = 0,00109 mol KOH

2. 0,00109 / 2 = 0,000545 mol H₂SO₄

3. 0,000545 / 0,3 = 0,0018 mol/l

b)

1. n = 0,02 * 0,0358 = 0,000716mol KOH

2. 0,000716 / 2 = 0,000358 mol H₂SO₄

3. 0,000358 / 0,3 = 0,0012 mol/l

c)

1. n= 0,02 * 0,0538 = 0,001076 mol KOH

2. 0,0010076 / 2 = 0,000538 mol H₂SO₄

3. 0,000538 / 0,3 = 0,00179 mol/l

d)

1. n = 0,02 * 0,0541 = 0,001082 mol KOH

2. 0,001082 / 2 = 0,000541 mol H₂SO₄

3. 0,000541 / 0,3 = 0,0018 mol/l

To find of the concentration of H2SO4 in the lake, we find the average value of each trials result:

(0,0018+0,0012+0,00179+0,0018) / 4 = 0,0016475 mol/l

ANSWER: 0,0016475 mol/l H₂SO₄