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What is the pH of a solution containing 0 250 M HONH_2 and 0.400 M HONH_3CI? (K_

ID: 1048890 • Letter: W

Question

What is the pH of a solution containing 0 250 M HONH_2 and 0.400 M HONH_3CI? (K_b for HONH_2 is 1.1 times 10^-8) 7.75 8.16 5.84 6.25 12 73 A 150.0 mL buffer solution containing 0.100 M hydrocyanic acid (HCN. K_a = 6.2 times 10^-10) and 0 100 M potassium cyanide (KCN) has 30.0 mL of 0.120 M KOH added to it. What is the pH of the solution after the KOH has been added? 8 73 9.00 9 21 9.29 9.42 The K_a of acetic acid is 1 76 times 10^-5 The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium acetate and 50.0 mL of 1.00 M acetic acid is 0.85 1.70 2.38 3.40 4.75

Explanation / Answer

15. [HONH2] base = 0.25 M

    [HONH3Cl] salt = 0.4 M

   pkb = -logkb

       = -log(1.1*10^-8)

       = 8
pH of basic buffer = 14 - [pkb+log(salt/base)]

pH = 14-(8+log(0.4/0.25))

      = 5.8

answer: C.5.84

16. no of mole of HCN added = 150*0.1 = 15 mmole

    no of mole of KCN added = 150*0.1 = 15 mmole

no of mole of KOH = 30*0.12 = 3.6 mmole

pH of buffer = pka + log(salt+KOH/acid-KOH)

         pka = -logka

             = -log(6.2*10^-10)

            =   9.2

   pH = 9.2+log((150+3.6)/(150-3.6))

      = 9.22

answer :C.9.21

17. from the data

   no of mole of CH3COOK = 50*1 = 50 mmole

no of mole of CH3COOH = 50*1 = 50 mmole

   pka = -logKa

        = -log(1.76*10^-5)

        = 4.75

as no of mole of CH3COOK = no of mole of CH3COOH

   pH = pka = 4.75

answerr: e.4.75

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