1.) A sample of 20.0mL sodium carbonate of 0.05M was titrated against hydrochlor

Question

1.) A sample of 20.0mL sodium carbonate of 0.05M was titrated against hydrochloric acid was taken in a burette. The color of the indicator changed when the burette reading read 25.5mL. What was the concentration of the hydrochloric acid taken in the burrette? Please show all work, and write the complete balanced chemical reaction involved in the process.

2.) Calculate the freezing point of a solution containing 2.5g of Potassium nitrate and 575mL of water. Assume that the density of water at the temperature is 0.998g/cc. The molal freezing point depression constant (Kf) for water is .86 degrees C/m



1.) A sample of 20.0mL sodium carbonate of 0.05M was titrated against hydrochloric acid was taken in a burette. The color of the indicator changed when the burette reading read 25.5mL. What was the concentration of the hydrochloric acid taken in the burrette? Please show all work, and write the complete balanced chemical reaction involved in the process.

2.) Calculate the freezing point of a solution containing 2.5g of Potassium nitrate and 575mL of water. Assume that the density of water at the temperature is 0.998g/cc. The molal freezing point depression constant (Kf) for water is .86 degrees C/m





2.) Calculate the freezing point of a solution containing 2.5g of Potassium nitrate and 575mL of water. Assume that the density of water at the temperature is 0.998g/cc. The molal freezing point depression constant (Kf) for water is .86 degrees C/m

Explanation / Answer

mmol of base =MV = 20*0.05 = 1 mmol of base

ratio is 2:1 so

1 mmol obase = 2 mmol of acid

then

[acid] = mmol/V = 2/25.5 = 0.07843 M

2-

m = 2.5 KNO3

mol = 2.5/101.1032 = 0.024727

NOTE --> actual depression value for water is 1.86 not only 0.86

dTf = -Kf*m*i

i = 2 ions so = 2

dTf = -1.86*(0.024727)/(0.575) * 2

dTf = -0.1599729

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