Answer to Question #167645 in General Chemistry for Marifel Colina

Question #167645

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.



1
Expert's answer
2021-03-01T02:36:12-0500

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) <==> HPO42- (aq) + H+ (aq) pKa2 = 7.21

HPO42- (aq) <==> PO43- (aq) + H+ (aq) pKa3 = 12.3



in the question, we are provided the concentration of HPO42- and PO43-.


So we are interested in 3rd dissociation of H3PO4.


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


HPO42- will be the acid, and PO43- 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, [HPO42-]= 0.33M and [PO43-] = 0.52 M , we have


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



"pH = 12.497 ;"


In 2 decimal place, the answer is 12.50


Hence the pH of the HPO42-/PO43-. buffer is 12.50.


----------------------------------------x-------------------------------------------------------x-------------------------------------------------------


 (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.



------------------------------x-------------------------------------------x-------------------------------------


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

Solution:


H2PO4- / HPO42- would be our buffer system.

In problem a, we say the reaction.


H2PO4- (aq) <==> HPO42- (aq) + H+ (aq) pKa2 = 7.21


H2PO4- will the acid, and HPO42- will be its conjugate base.


We are now given the pH of buffer = 7.65,

and asked to find the ratio of H2PO4- / HPO42- 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

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