B. H solution for neutralization of HCl (aq) and NaOH (aq) : Volume of HCl: 0.05

Question

B. Hsolution for neutralization of HCl(aq) and NaOH(aq):
Volume of HCl: 0.0500 L
Volume of NaOH: 0.0500 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Molarity of NaOH: 1.95 M
T for reaction B: 11.0°C

1.Calculate the molarity of the resulting sodium chloride solution.

2.Calculate the value (calories) for the heat of reaction.

3.Calculate H for neutralization.

BACKGROUND INFO:

SPECIFIC HEAT OF WATER= 1.00 DENSITY=1.00

SPECIFIC HEAT OF SODIUM HYDROXIDE= 0.95 DENSITY=1.04

SPECIFIC HEAT OF SODIUM CHLORIDE=0.94 DENSIT=1.04

SPECIFIC HEAT OF POTASSIUM CHLORIDE= 0.90 DENSITY=1.10

Explanation / Answer

1) The molarity of HCl = Molarity of NaOH

Moles of HCl = Moles of NaOH = Volume X molarity = 1.95 X 0.05 = 0.0975 moles

So moles of NaCl formed = 0.0975 moles

So molarity = moles of NaCl / total volume of solution = 0.0975 / 0.05 + 0.05 = 0.975 Molar

2) Heat = mass x T change x specific heat

Mass of water = 100 grams

Heat = 100 X 11 X 1 cal / g = 1100 calories

3)Gram equivalents of acid base neutralized = 0.0975

We know that the heat of neutralization for strong acid and base = 13.7 Kcal /g eqwt

so for 0.0975 gram equivalent heat of neutralizaiton will be = 13.7 X 0.0975 = 1.33 Kcal

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