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Snezhnost [94]
4 years ago
9

A 1 liter solution contains 0.247 M nitrous acid and 0.329 M sodium nitrite. Addition of 0.271 moles of calcium hydroxide will:

(Assume that the volume does not change upon the addition of calcium hydroxide.)
a. Raise the pH slightly
b. Lower the pH slightly
c. Raise the pH by several units
d. Lower the pH by several units
e. Not change the pH
f. Exceed the buffer capacity
Chemistry
1 answer:
Inessa05 [86]4 years ago
4 0

Answer:

a. Raise the pH slightly

Explanation:

We know that

Pka of HNO2/KNO2 =3.39

Moles of HNO2 in the buffer=0.247 mol/L×1L=0.247 moles

Moles of NO2-=0.329mol/L×1L=0.329 moles

If 0.271 moles of Ca(OH)2 is added it will neutralise 0.136 moles of acid ,HNO2,remaining HNO2=0.247-0.136=0.111 moles

Moles of NO2- will increase as 0.0333 moles Ca(NO)2 will be formed =0.0333+0.036=0.0693 moles

pH=pka+log [base]/[acid]      {henderson -hasselbach equation}

=3.39+log (0.0693/0.0317)=3.39+0.34=3.73

pH=3.73

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Explanation:

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An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/ml; = 62.07 g/mol) and water (d
tankabanditka [31]
a) Volume percent

Formula: % v/v = [volume solute / volume solution] * 100

Just to make it easy take a base of 50 volume parts of ethylen glycol and 50 volume parts of water to make 100 volumes of mixture (this assumpion will be valid for all the questions):

% v/v =[ 50 ml ethyleneglycol] / [100 ml mixture] * 100 = 50%

Answer: 50% v/v

b) Mass percent

% m/m = [mass ethylene glycol / mass solution] * 100

mass ethylene glycol = 50 ml * 1.114 g/ ml = 55.7 g

mass of mixture = 100 ml * 1.07 g/ml = 107 g

% m/m = [55.7 / 107 g] * 100 = 52.06 %

Answer: 52.06%

c) Molarity

M = number of moles of solute / liters of solution

number of moles of solute = mass in grams / molar mass

number of moles of ehtylene glycol = 55.7 g / 62.07 g/mol = 0.8974 mol

liters of solution = 0.1 liter

M = 0.8974 mol / 0.1 liter = 8.974 M

Answer: 8.974 M

d) Molality

m = number of moles of solute / kg of solvent

number of moles of ethylen glycol = 0.8974 mol

mass of water = 50 ml * 1 g/ml = 50 g = 0.05 kg

m = 0.8974 mol / 0.05 kg = 17.95 m

Answer: 17.95 m

e) mole fraction

mole fraction = [number of moles of solute] / [number of moles of mixture] * 100

number of moles of ethylen glycol = 0.8974 mol

number of moles of water = 50 g / 18.01 g /mol = 2.776 mol

mole fraction = 0.8974 mol / [0.8974 mol + 2.776 mol] = 0.244

Answer: 0.244
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3 years ago
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