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o-na [289]
3 years ago
10

Calculate the pH of a solution that is 0.205M benzoic acid and 0.230M sodium benzoate, a salt whose anion is the conjugate base

of benzoic acid. (The pKa of benzoic acid
Chemistry
1 answer:
pickupchik [31]3 years ago
4 0

Answer:

pH = 4.25

Explanation:

A solution composed of a weak acid and its conjugate base is a <em>buffer solution</em>. To calculate the pH of a buffer solution we use the Henderson-Hasselbach equation:

pH = pKa + log ([conjugate base]/[weak acid]

In this case, we have the following data:

[conjugate base] = [sodium benzoate] = 0.230 M

[weak acid] = [benzoic acid] = 0.205 M

The pKa of benzoic acid is 4.2. So, we introduce the data in the equation:

pH = 4.2 + log (0.230 M/0.205 M) = 4.2 + 0.050 = 4.25

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Explanation :

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6 0
3 years ago
The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
Agata [3.3K]

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

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