Question

A sample of 0.7960 g of an unknown compound containing barium ions (Ba2+) is dissolved in water and treated with an excess of Na2SO4. If the mass of the BaSO4 precipitate formed is 0.7651 g, what is the percent by mass of Ba in the original unknown compound?

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How many grams of NaCl are required to precipitate most of the Ag+ ions from 2.50 × 102 mL of 0.0467 M AgNO3 solution

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How many grams of KHP are needed to neutralize 75.47 mL of a 0.1041 M NaOH solution?

(molar mass of KHP = 204.2 g/mo

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How many grams of KHP are needed to neutralize 75.47 mL of a 0.1041 M NaOH solution?

(molar mass of KHP = 204.2 g/mo

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Calculate the volume of a 1.420 M NaOH solution required to titrate 36.75 mL of a 1.500 M H3PO4 solution.

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What volume of a 0.500 M HCl solution is needed to neutralize each of the following:

(a) 16.0 mL of a 0.300 M NaOH solution

(b) 10.0 mL of a 0.200 M Ba(OH)2 solution

Answer 1

1)
Ba+2 + Na2SO4 ------> BaSO4 + 2 Na+2

1 mole of BaSO4 require 1 mole of Ba+2

233.38 g/mol of BaSO4 require 137.3 g of Ba+2

0.7651 g of BaSO4 requires 137.3/233.38 * 0.7651 g of Ba+2

                          = 0.4501 g

percentage of Ba+2 = 0.4501/0.796 * 100 = 56.55 %

2)

number of moles of AgNO3 = 2.5 * 10^2 *10^-3* 0.0467 = 0.01167 moles

NaCl + AgNO3 ------> AgCl + NaNO3

1 mole of AgNoe requires 1 mole of NaCl

0.01167 moles of AgNO3 requires 0.01167 moles of NaCl

mass of NaCl = 0.01167 * 58.5 = 0.6827 g

3)

KHP + NaOH ------> NaKHP + H2O

1 mole of NaOH requires 1 mole of KHP

75.47 * 10^-3 * 0.1041 moles of NaOH requires 75.47 * 10^-3 * 0.1041 moles of KHP

mass of KHP = 75.47 * 10^-3 * 0.1041 * 204.22 = 1.6044 g

4)

H3PO4 + 3NaOH ------> Na3PO4 + 3H2O

M1V1/n1 = M2V2/n2

1.5 * V1/1 = 1.42 * 36.75/3

V1 = volume of H3PO4 solution = 11.6 mL

5)

NaOH + HCl ----> NaCl + H2O

M1V1/n1 = M2V2/n2

0.3 * 16/1 = 0.5 * V2/1

Volume of HCl V2 = 9.6 mL

Ba(OH)2 + 2HCl ------> BaCl2 + 2H2O

M1V1/n1 = M2V2/n2

0.2 * 10/1 = 0.5 * V2/2

Volume of HCl V2 = 8 mL