# How many milliliters of 2.5M benzoic acid need to be added to 500. mL of a solution already contains 6.00 grams of sodium benzoate, if you want the final pH to be 4.55? Ka for benzoic acid =6.28x10^-5 (B)if you add 16 grams of sodium benzoate to 4.55 grams of benzoic acid to a 1.00 liter container of water, what is the pH of this solution? (c)how many grams of potassium nitrite need to be added to 250.0 mL of a 0.1500 M HNO2 solution to produce a final buffer at pH=3.65? Ka for HNO2=4.5x10^-4

Question
Asked Nov 25, 2019
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How many milliliters of 2.5M benzoic acid need to be added to 500. mL of a solution already contains 6.00 grams of sodium benzoate, if you want the final pH to be 4.55? Ka for benzoic acid =6.28x10^-5

(B)if you add 16 grams of sodium benzoate to 4.55 grams of benzoic acid to a 1.00 liter container of water, what is the pH of this solution?

(c)how many grams of potassium nitrite need to be added to 250.0 mL of a 0.1500 M HNO2 solution to produce a final buffer at pH=3.65? Ka for HNO2=4.5x10^-4

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Step 1

The value of pKa is calculated below.

pKa = -logKa

= - log (6.28 × 10-5)

= 4.2

The ratio of sodium benzoate to benzoic acid is calculated in equation (1) with the help of Henderson-Hasselbalch equation.

…… (1)

On solving it further as shown in equation (2),

Step 2

The molar mass of sodium benzoate is 144.11 g/mol. The number of moles (n) of sodium benzoate is calculated in equation (3).

…… (3)

The molarity of sodium benzoate is calculated in equation (4) by substituting 500 mL forV and 0.0416 mol for n.

Step 3

The initial concentration (M1) of Benzoic acid is 2.5 M. The volume (V2) after the addition of two solution becomes x L + 0.5 L. Thus, the final concentration (M2) of Benzoic acid is given in equation (5).

The initial concentration (M1) of sodium benzoa...

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