Calculate [H*] and [OH] for solutions with the following pH values 4.0 8.52 12.6

Question

Calculate [H*] and [OH] for solutions with the following pH values 4.0 8.52 12.60 2 30 mL of a 0.4375 M HCI solution are added to 35 mL of 0.250 NaOH solution. What is the final pH? A solution with a pH of 2.91 is prepared by dissolving 1 g. ofbcnzoic acid, HC_2H_2O_2, a weak acid. in 350 ml. of water. What is K_a of benzoic aad? A buffer is prepared in which the ratio [HCO_3YV[CO,2] is 4 0 What is the pH of this buffer (K_a HCO_3 = 4.7 x 10^-11 )? What is the pH of the buffer if 0.0100 moles of Hcl are added? What is the pH of the buffer if 0.0200 moles of NaOH are added?

Explanation / Answer

1.PH = 4
-log[H+] = 4
     [H+] = 10^-4 M
     Kw = [H+][OH-]
     1*10^-14 = 10^-4[OH-]
     [OH-] =10^-10M
b. PH = 8.52
-log[H+] = 8.52
     [H+] = 10^-8.52 = 3.091*10^-9
Kw = [H+][OH-]
     1*10^-14 = 3.091*10^-9[OH-]
      [OH-] = 3.23*10^-5M

c. PH = 12.6
-log[H+] = 12.6
   [H+] = 10^-12.6
   [H+] = 2.51*10^-13M
Kw = [H+][OH-]
     1*10^-14 =2.51* 10^-13[OH-]
    [OH-] = 3.98*10^-2 M
2. M = MBVB-MAVA/VA+VB
      = 0.25*35-0.4375*30/35+30
      = 0.067M
    [OH-] = 0.067M
   POH = -log[OH-]
      = -log0.067
      = 1.1739
   PH = 12.8261
3. m = 1*1000/122*350 = 1000/42700 = 0.0234M

    PH = 2.91
   -log[H+] = 2.91
      [H+] = 10^-2.91 = 0.00123M
     [H+] = squaroot of Ka*c
     0.0012 =squaroot of Ka*0.023
    0.00000144 = Ka*0.023
   Ka = 6.26*10^-5

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