1) A student is asked to standardize a solution of calcium hydroxide. He weighs

Question

1) A student is asked to standardize a solution of calcium hydroxide. He weighs out 1.01 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 27.2 mL of calcium hydroxide to reach the endpoint.

A. What is the molarity of the calcium hydroxide solution? M

This calcium hydroxide solution is then used to titrate an unknown solution of hydrobromic acid.

B. If 23.3 mL of the calcium hydroxide solution is required to neutralize 13.9 mL of hydrobromic acid, what is the molarity of the hydrobromic acid solution? M

2) A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.01 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).
It requires 22.3 mL of potassium hydroxide to reach the endpoint.

A. What is the molarity of the potassium hydroxide solution?  M

This potassium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid.

B. If 17.9 mL of the potassium hydroxide solution is required to neutralize 21.3 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution?  M

Explanation / Answer

Solution:- (1) (A)

mass of KHC8H4O4 is given = 1.01 g

volume of Ca(OH)2 used = 27.2 ml = 0.0272 L

The balanced equation is....

2KHC8H4O4 + Ca(OH)2 -----------> 2H2O + Ca(KC8H4O4)2

From balanced equation there is 2:1 mol ratio between KHC8H4O4 and Ca(OH)2.

Let's convert grams of KHC8H4O4 into moles and then calculate the moles of Ca(OH)2 by using the mol ratio.

molar mass of KHC8H4O4 is 204.44 g/mol

1.01 g KHC8H4O4 x (1mol/204.44 g) x (1 mol Ca(OH)2/2 mol KHC8H4O4) = 0.00247 mol Ca(OH)2

molarity = mol/L

so, molarity of Ca(OH)2 = 0.00247 mol/0.0272 L = 0.0908 M

(B) Now this solution of Ca(OH)2 is used to titrate HBr. Balanced equation is...

2HBr + Ca(OH)2 ----------> 2H2O + CaBr2

23.3 ml that is 0.0233 L of Ca(OH)2 are used. molarity of this is 0.0908 M. so, moles of Ca(OH)2 used to titrate HBr = 0.0233 L x 0.0908 mol/L = 0.00212 mol

there is 1:2 mol ratio between Ca(OH)2 and HBr. So, moles of HBr could be calculated as...

0.00212 mol Ca(OH)2 x 2 mol HBr/1 mol Ca(OH)2 = 0.00424 mol HBr

volume of HBr used = 13.9 ml = 0.0139L

molarity of HBr = 0.00424 mol/0.0139 L = 0.305 M

(2) (A) The balanced equation is....

KHC8H4O4 + KOH ---------> H2O + K(KC8H4O4)

let's calculate the moles of KOH same as we calculated the moles of Ca(OH)2 for first problem.

1.01 g KHC8H4O4 x (1mol/204.44 g) x (1 mol KOH/1 mol KHC8H4O4) = 0.00494 mol KOH

Volume of KOH used = 22.3 ml = 0.0223 L

molarity of KOH = 0.00494 mol/0.0223L = 0.222 M

(B) KOH + HI ----------> H2O + KI

They react in 1:1 mol ratio.

Volume of KOH used = 17.9 ml = 0.0179 L

moles of KOH used = 0.0179 L x 0.222 mol/L = 0.00397 mol

moles of HI would also be 0.00397 since they react in 1:1 mol ratio.

volume of HI used = 21.3 ml = 0.0213 L

molarity of HI = 0.00397 mol/ 0.0213 L = 0.186 M

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