Answer to Question #167645 in General Chemistry for Marifel Colina

Question

Calculate the pH of buffers that contain the acid and conjugate base in following concentrations:

 (a) [HPO42-] = 0.33 M, [PO43-] = 0.52 M

 (b) [CH3COOH] = 0.40 M, [CH3COO-] = 0.25

What ratio of concentrations of NaH2PO4 and Na2HPO4 in solution would give a buffer with pH = 7.65?

1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq) (Kb = 1.8 x 10-5 mol.L-1) . Calculate the pH of the solution.

Answer 1

Q167645

Calculate the pH of buffers that contain the acid and conjugate base in following concentrations:

 (a) [HPO42-] = 0.33 M, [PO43-] = 0.52 M

 (b) [CH3COOH] = 0.40 M, [CH3COO-] = 0.25

What ratio of concentrations of NaH2PO4 and Na2HPO4 in solution would give a buffer with pH = 7.65?

1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq) (Kb = 1.8 x 10-5 mol.L-1).

Solution:

a) H₃PO₄ is a tri-protic acid with pKa values.

H₃PO₄(aq) <==> H₂PO₄^- (aq) + H⁺ (aq) , pKa ₁ = 2.12

H₂PO₄^- (aq) <==> HPO₄^2- (aq) + H⁺ (aq) pKa₂ = 7.21

HPO₄^2- (aq) <==> PO₄^3- (aq) + H⁺ (aq) pKa₃ = 12.3

in the question, we are provided the concentration of HPO₄^2- and PO₄^3-.

So we are interested in 3rd dissociation of H₃PO₄.

We will use the Henderson-Hasselbalch equation for finding the pH of the given buffer.

HPO₄^2- will be the acid, and PO₄^3- will be its conjugate base.

Henderson-Hasselbalch equation

$pH = pKa + log \frac{[A^-]}{[HA]} = pKa + log \frac{[Base]}{[Acid]}$

$pH = pKa + log \frac{[PO_4^{3-}]}{[HPO_4^ {2-}]}$

plug, pKa = 12.3, [HPO₄^2-]= 0.33M and [PO₄^3-] = 0.52 M , we have

$pH = 12.3 + log \frac{0.52M}{0.33M} = 12.3 + log[1.576 ] = 12.3 + 0.1975 pH = 12.497 ;$

$pH = 12.497 ;$

In 2 decimal place, the answer is 12.50

Hence the pH of the HPO₄^2-/PO₄^3-. buffer is 12.50.

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 (b) [CH₃COOH] = 0.40 M, [CH₃COO-] = 0.25.

This problem would be solved in a similar way.

Acetic acid is a monoprotic acid with pKa = 4.745, Sometimes instead of pKa, you may be given Ka.

Ka of tactic acid is 1.8 * 10^-5.

If you are provided the Ka then use the formula

pKa = - log [Ka] and find the pKa first

pKa = - log [ 1.8 * 10^-5 ] = 4.745 ;

CH₃COOH is the acid and CH₃COO-is its conjugate base.

Henderson-Hasselbalch equation

$pH = pKa + log \frac{[A^-]}{[HA]} = pKa + log \frac{[Base]}{[Acid]}$

$pH = pKa + log \frac{[CH_3COO^-]}{[CH_3COOH]}$

plug, pKa = 4.745 , [CH₃COOH]= 0.40M and [CH₃COO^- ] = 0.25 M , we have

$pH = 4.745 + log \frac{0.25}{0.40M} = 4.745 + log[0.625 ] = 4.745 - 0.2041$

$pH = 4.541 ;$

In 1 decimal place, the answer is 4.54

Hence the pH of the CH₃COOH/ CH₃COO- buffer is 4.54.

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What ratio of concentrations of NaH2PO4 and Na2HPO4 in solution would give a buffer with pH = 7.65?

Solution:

H₂PO₄^- / HPO₄^2- would be our buffer system.

In problem a, we say the reaction.

H₂PO₄^- (aq) <==> HPO₄^2- (aq) + H⁺ (aq) pKa₂ = 7.21

H₂PO₄^- will the acid, and HPO₄^2- will be its conjugate base.

We are now given the pH of buffer = 7.65,

and asked to find the ratio of H₂PO₄^- / HPO₄^2- required for having pH = 7.65.

We will put pKa2 = 7.21, and pH = 7.65 , and find the ratio

Henderson-Hasselbalch equation

$pH = pKa + log \frac{[A^-]}{[HA]} = pKa + log \frac{[Base]}{[Acid]}$

$pH = pKa + log \frac{[HPO_4^{2-}]}{[H_2PO_4^ {-}]}$

$7.65 = 7.21 + log \frac{[HPO_4^{2-}]}{[H_2PO_4^ {-}]}$

subtract 7.21 from both the side we have

$0.44 = log \frac{[HPO_4^{2-}]}{[H_2PO_4^ {-}]}$

taking 10^ on both the side we have

$10^{0.44} = 10^{ log \frac{[HPO_4^{2-}]}{[H_2PO_4^ {-}]}}$

$\frac{2.754 }{1} = \frac{[HPO_4^{2-}]}{[H_2PO_4^ {-}]}$

in the question, we are asked the ratio of concentrations of NaH2PO4 and Na2HPO4.

which means we are asked the ratio H2PO4^- / HPO4^2-

hence we will invert both sides of the equation.

flipping both sides of the equation we have

$\frac{1 }{2.754} = \frac{[H_2PO_4^ {-}]}{[HPO_4^{2-}]}$

$0.3631 = \frac{[H_2PO_4^ {-}]}{[HPO_4^{2-}]}$

Hence the ratio of concentrations of NaH2PO4 and Na2HPO4 should be 0.3631/ 1