Question

24. How many milliliters of 2.00 M H,SO, will be required to neutralize 45.0 mL of 3.00 N KOH? 25. A careless laboratory assistant set out to determine the concentration of a perchloric acid (HCIO.) solution. He found that 50.00 mL of 1.000 N NaOH neutralized 0.5000 L of the HClO, solution. His calculations were as follows: N (50.00 mL)(1.000 N) 2 = 100.0 N HCIO, 0.5000 L Another laboratory assistant insists that the calculation is incorrect. Why? What is the correct answer? 26. A salad dressing manufacturer desires to make tangy salad dressing by using vinegar with an acetic acid concentration of at least 1 N. The quality control department examines a sample of vinegar and determines that 300.0 mL of 0.100 N sodium hydroxide neutralizes the acetic acid in 25.00 mL of the vinegar. Does this vinegar meet the man- ufacturer's requirements?

Answer 1

24. At the equivalence point in the neutralization, the moles of acid is equal to moles of base

Moles of acid = Moles of base

This can also be written as

M_A × V_A = M_B × V_B ⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱(1)                      

Where M_A and M_B indicates the molarity of acid and base while V_A andV_B are their volumes

Here the neutralization reaction can be given as

H₂SO₄ (aq) + 2 KOH (aq) K₂SO₄ (aq)+ 2 H₂O (l) ⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱(2)                      

Here 2 moles of KOH are required to neutralize 1 mole of H₂SO₄

Given molarity of acid, H₂SO₄ = 2.00 M

Volume of base, KOH = 45.0 mL

Normality of base, KOH = 3.00 N

In order to use the equation (1) we have to convert normality of KOH into molarity

Now, Molarity = Normality × (Molecular weight/equivalent weight)

For KOH molecular weight = equivalent weight, therefore molarity = normality = 3.00 M.

For (2) at equivalence point,

Moles of KOH = Vol. of KOH × Molarity of KOH

Moles of KOH = 0.045 L × 3 M = 0.135 moles of KOH

Now,

0.135 mol of KOH × = 2.00 M H₂SO₄ × (x L of H₂SO₄)

1 mol H2SO4 2 mol KOH

x L of H₂SO_{4 ­}= 0.0675 mol/2 M = 0.03375 L

Volume of H₂SO₄ at neutralization point in mL = 33.75 mL

25. The reaction between strong acid HClO₄ and a strong base NaOH can be written as

HClO₄ (aq) + NaOH (aq) NaClO₄ (aq)+ H₂O (l) ⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱(3)       

Given

The normality = molarity of NaOH = 1.00 M

Volume of NaOH used for neutralization = 50.0 mL = 0.050 L

Volume of HClO₄ at neutralization = 0.500 L

The equation given in the question is incorrect since the volume of the base used at the neutralization point is given in mL whereas care must be taken to give the volume of the acid/base used at the neutralization point in liter (L) only.

Therefore, the equation will now be given as

N or M = (0.050 L) × = 0.1 M

(0.050 L) X (1.00 M) 0.500 L

Please note: Here we can use molarity and normality interchangeably since the equivalent and molecular weight of both the species are the same.

26. A condition is given in salad dressing that the concentration of vinegar should be at least 1 N.

Given

Volume of NaOH at neutralization = 300 mL = 0.30 L

Normality of NaOH = 0.100 N

Volume of Vinegar (CH₃COOH) = 25.00 mL = 0.025 L

Now a neutralization reaction is performed which has following equation

CH₃COOH (aq) + NaOH (aq) CH₃COONa (aq)+ H₂O (l) ⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱(4)  

So at neutralization point

Moles of CH₃COOH = Moles of NaOH⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱(5)  

Now, Moles of NaOH = Normality of NaOH × Volume of NaOH

Moles of NaOH = 0.1 N × 0.30 L = 0.03 mol NaOH

Moles of CH₃COOH = Normality of CH₃COOH × Volume of CH₃COOH

Moles of CH₃COOH = N × 0.025 L

Now substituting these values in equation (5)

N × 0.025 L = 0.03 mol NaOH

Normality of CH₃COOH = 1.2 N

Since the normality of vinegar made is above 1 N it meets the requirements of the manufacturer.