Answer to Question #165730 in General Chemistry for Anthony

Question #165730

A 100.00 mL buffer solution is prepared by dissolving 1.22 grams benzoic acid (pKa = 4.20, Molar Mass = 122g/mol) and 2.88 grams sodium benzoate (Molar Mass = 144g/mol) in an appropriate amount of water.

a. What is the pH of the buffer solution?

b. What is the pH of the solution if 10.00 mL of 0.25 M HCl is added?

c. What is the pH of the solution if 5.00 mL of 0.25 M NaOH is added instead of HCl?


1
Expert's answer
2021-02-23T04:26:33-0500

a.)

We have given,

Number of moles of "C_6H_5COOH" , "n_1" ="\\dfrac{1.22}{122} = 0.01"


Number of moles of"C_6H_5COONa,n_2" ="\\dfrac{2.88}{144} = 0.02"


Concentration of "C_6H_5COOH, M_1" = 0.1M


Concentration of "C_6H_5COONa, M_2" = 0.2M


We know, pH in a buffer solution can be calculated as

"pH = pK_a+log\\dfrac{M_2}{M_1}\\\\\n\npH = pK_a+log\\dfrac{0.2}{0.1}\\\\\n\n\n\n\npH = 4.20+0.30\\\\\n\npH = 4.50"

b.)

Now, 10.00 mL of 0.25 M "HCl" is added then,


"M_1V_1 = M_2V_2"


"0.25\\times10 = M2\\times100"


"M_2 = 0.025M"


where "M_2" is the concentration of solution when "HCl" is added

then,

"pH = -log[H]^+"


"pH = -log[0.025]"


"pH = 1.60"

c.)

When 5.00 mL of 0.25 M "NaOH" is added then,


"M_1V_1 = M_2V_2"


"0.25\\times5 = M2\\times100"


"M2 = 0.0125M"


"pH = -log[H]^+"


"pH = 1.90"


Hence, pOH = "14-1.90"

= "12.1"


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