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Anuta_ua [19.1K]
3 years ago
9

How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the p

H to 5.05 ?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
8 0
Initial moles of CH₃COOH = 1.00 x 0.0100 = 0.0100 mol

Initial moles of CH₃COONa = 1.00 x 0.100 = 0.100 mol


Let a be the volume of HNO₃ that must be added

Moles of HNO₃ added = a/1000 x 10.0 = 0.01a mol

CH₃COONa + HNO₃ => CH₃COOH + NaNO₃


Moles of CH₃COOH = 0.0100 + 0.01a

Moles of CH₃COONa = 0.100 - 0.01a


pH = pKa + log([CH₃COONa]/[CH₃COOH])

= pKa + log(moles of CH₃COONa/moles of CH₃COOH)


5.05 = 4.75 + log((0.100 - 0.01a)/(0.0100 + 0.01a))

log((0.100 - 0.01a)/(0.0100 + 0.01a)) = 0.3

(0.100 - 0.01a)/(0.0100 + 0.01a) = 10^0.3 = 2.0

0.03 a = 0.08

a = 2.67


<span>Volume of HNO3 = a = </span>2.13 mL
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Answer:

The molality of the KCl solution is 11.8 molal

Explanation:

Step 1: Data given

Mol fracrion KCl = 0.175

Molar mass KCl = 74.55 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mol fraction H2O

mol fraction H2O = 1 - 0.175 = 0.825

Step 3: Calulate mass of H2O

Suppose the total moles = 1.0 mol

Mass H2O = moles H2O * molar mass

Mass H2O = 0.825 * 18.02 g/mol

Mass H2O = 14.87 grams = 0.01487 kg

Step 4: Calculate molality

Molality KCl = 0.175 / 0.01487 kg

Molality KCl = 11.8 molal

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