57.8 mL of 0.993 M hydroiodic acid is added to 37.4 mL of calcium hydroxide, and

Question

57.8 mL of 0.993 M hydroiodic acid is added to 37.4 mL of calcium hydroxide, and the resulting solution is found to be acidic. 15.4 mL of 2.61 M potassium hydroxide is required to reach neutrality. What is the molarity of the original calcium hydroxide solution?

40.5 mL of 1.46 M nitric acid is added to 34.3 mL of sodium hydroxide, and the resulting solution is found to be acidic. 27.8 mL of 0.669 M calcium hydroxide is required to reach neutrality. What is the molarity of the original sodium hydroxide solution?

Explanation / Answer

Q1

HI = 0.993 M, 57.8 mL

Ca(OH)2 = X; 37.4 mL

solution is sitll acidic

then

KOH = MV = 15.4 * 2.61 = 40.194 mmol of KOH

mmol of HI = MV = 57.8*0.993 = 57.3954 mmol of H+

then

mmol of OH- used in Ca(OH)2 = (57.3954-40.194 ) = 17.2014 mmol of OH-

mmol of Ca(OH)2 = 1/2*17.2014 = 8.6007 mmol of Ca(OH)2

[Ca(OH)2] = mmol/V = 8.6007 / 37.4 = 0.2299 M

Q2.

mmol of HNO3 = MV = 40.5*1.46 = 59.13

mmol of NaOH = MV = 34.3*X =

mmol of Ca(OH)2 = MV = 0.669*27.8 = 18.5982

mmol of OH- = 18.5982*2 = 37.1964

mmol of OH- usedas NaOH = 59.13-37.1964 = 21.9336

mmol of NaOH = MV = 34.3*X = 21.9336

X = 21.9336/34.3

[NaOH] = 0.6394 M

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