1A) 53.5 mL of 1.11 M nitric acid is added to 30.6 mL of sodium hydroxide , and

Question

1A) 53.5 mL of 1.11 M nitric acid is added to 30.6 mL of sodium hydroxide, and the resulting solution is found to be acidic.
21.4 mL of 0.659 M barium hydroxide is required to reach neutrality.
What is the molarity of the original sodium hydroxide solution?

1B)47.3 mL of 0.249 M hydrobromic acid is added to 35.5 mL of barium hydroxide, and the resulting solution is found to be acidic.
28.0 mL of 0.189 M potassium hydroxide is required to reach neutrality.
What is the molarity of the original barium hydroxide solution?

1C)Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP.
How many grams of KHP are needed to exactly neutralize 34.2 mL of a 0.539 M barium hydroxide solution ?

Explanation / Answer

1A) Write the balanced chemical equations.

HNO3 (aq) + NaOH (aq) -------> NaNO3 (aq) + H2O (l)

2 HNO3 (aq) + Ba(OH)2 (aq) -------> Ba(NO3)2 (aq) + 2 H2O (l)

As per the stoichiometric equations,

1 mole HNO3 = 1 mole NaOH

2 moles HNO3 = 1 mole Ba(OH)2

Millimole(s) of HNO3 added = (53.5 mL)*(1.11 M) = 59.385 mmole.

Millimole(s) of Ba(OH)2 added = (21.4 mL)*(0.659 M) = 14.1026 mmole.

Millimoles of HNO3 reacted with 14.1026 mmole Ba(OH)2 = (14.1026 mmole Ba(OH)2)*(2 moles HNO3/1 mole Ba(OH)2) = 28.2052 mmole.

Millimoles of HNO3 reacted with NaOH = (59.385 – 28.2052) mmole = 30.1798 mmole = millimoles of NaOH neutralized.

Molarity of NaOH = (millimoles of NaOH neutralized by HNO3)/(volume of NaOH in mL) = (30.1798 mmole)/(30.6 mL) = 0.9863 M ? 0.986 M (ans).

1B) Write the balanced chemical equations.

2 HBr (aq) + Ba(OH)2 (aq) -------> BaBr2 (aq) + 2 H2O (l)

HBr (aq) + KOH (aq) -------> KBr (aq) + H2O (l)

As per the stoichiometric equations,

2 moles HBr = 1 mole Ba(OH)2

1 mole HBr = 1 mole KOH

Millimole(s) of HBr added = (47.3 mL)*(0.249 M) = 11.7777 mmole.

Millimole(s) of KOH added = (28.0 mL)*(0.189 M) = 5.292 mmole.

Millimoles of HBr reacted with 5.292 mmole KOH = (5.292 mmole KOH)*(1 mole HBr/1 mole KOH) = 5.292 mmole.

Millimoles of HBr reacted with Ba(OH)2 = (11.7777 – 5.292) mmole = 6.4857 mmole

Millimoles of Ba(OH)2 neutralized = (6.4857 mmole HBr)*(1 mole Ba(OH)2/2 moles HBr) = 3.24285 mmole.

Molarity of Ba(OH)2 = (millimoles of Ba(OH)2 neutralized by HBr)/(volume of Ba(OH)2 in mL) = (3.24285 mmole)/(35.5 mL) = 0.0913 M ? 0.091 M (ans).

1C) Molar mass of KHC8H4O4 = (1*39.098 + 1*1.008 + 8*12.011 + 4*1.008 + 4*15.999) g/mol = 204.222 g/mol.

Write the balanced chemical equation for the reaction between KHP and Ba(OH)2.

Ba(OH)2 (aq) + 2 KHC8H4O4 (aq) -------> Ba(KC8H4O4)2 (aq) + 2 H2O (l)

As per the stoichiometric equation,

1 mole Ba(OH)2 = 2 moles KHC8H4O4.

Millimoles of Ba(OH)2 = (34.2 mL)*(0.539 M) = 18.4338 mmole.

Millimoles of KHP required to neutralize 18.4338 mmole Ba(OH)2 = (18.4338 mmole Ba(OH)2)*(2 moles KHP/1 mole Ba(OH)2) = 36.8676 mmole.

Mass of KHP required = (36.8676 mmole)*(1 mole/1000 mmole)*(204.222 g/1 mole) = 7.52917 g ? 7.529 g (ans).

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