An aqueous solution of hydrochloric acid is standardized by titration with a 0.1

Question

An aqueous solution of hydrochloric acid is standardized by titration with a 0.113 M solution of barium hydroxide.

If 22.8 mL of base are required to neutralize 21.4 mL of the acid, what is the molarity of the hydrochloric acid solution?

? M hydrochloric acid

What volume of a 0.270 M hydrochloric acid solution is required to neutralize 22.8 mL of a 0.113 M calcium hydroxide solution?

? mL hydrochloric acid

In the laboratory, a student dilutes 27.1 mL of a 9.21 M perchloric acid solution to a total volume of 150.0 mL. What is the concentration of the diluted solution?

Concentration = ? M

How many milliliters of 10.8 M hydrobromic acid solution should be used to prepare 2.50 L of 0.400 M ?

? mL

How many mL of 0.713 M HCl are needed to dissolve 9.29 g of BaCO3?

2HCl(aq) + BaCO3(s)   BaCl2(aq) + H2O(l) + CO2(g)

? mL

Explanation / Answer

1.

2 HCl (aq.) + Ba(OH)2 (aq.) --------> BaCl2 (aq.) + 2 H2O (l)

Neutralisation formula,

M1 V1 / n1 = M2 V2 / n2

M1 * 21.4 / 2 = 0.113 * 22.8 / 1

M1 = Molarity of acid = 0.241 M

2.

Neutralisation reaction:

2 HCl (aq.) + Ca(OH)2 (aq.) --------------> CaCl2 (aq.) + 2 H2O (l)

Neutrlisation formula,

M1 V1 / n1 = M2 V2 / n2

0.270 * V1 / 2 = 0.113 * 22.8 / 1

V1 = 19.1 mL

3.

Dilution formula,

M1 * V1 = M2 * V2

9.21 * 27.1 = M2 * 150.0

M2 = Final concentration = 1.66 M

4.

dilution formula,

M1 V1 = M2 V2

10.8 * V1 = 0.400 * 2.50

V1 = 0.0926 L

V1 = 92.6 mL

5.

Molar mass of BaCO3 = 197.3 g./mol

Moles of BaCO3 = mass / molar mass = 9.29 / 197.3 = 0.0471 mol

From the balanced equation,

1mol of BaCo3 needs 2 mol of HCl

Then,

0.0471 mol of BaCo3 needs 2 * 0.0471 = 0.0942 mol of HCl

Therefore,

Volume of HCl solution = moles / molarity = 0.0942 / 0.713 = 0.132 L = 132 mL

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