For the following, You will calculate the pH after the addition of different vol

Question

For the following, You will calculate the pH after the addition of different volumes of 0.3 M NaOH to a 0.1 M Nitrous acid solution

5. You have 100 mL of a 0.1 M HNO2 solution, and you add 25 mL of 0.3 M NaOH (aq). Calculate the pH. [Ka of HNO2 is 4.8 x 10-4]

6. You have 100 mL of a 0.1 M HNO2 solution, and you add 33.333 mL of 0.3 M NaOH (aq). Calculate the pH. [Ka of HNO2 is 4.8 x 10-4]

7. You have 100 mL of a 0.1 M HNO2 solution, and you add 50.0 mL of 0.3 M NaOH (aq). Calculate the pH. [Ka of HNO2 is 4.8 x 10-4]

Explanation / Answer

number of moles of HNO2 = 0.1 *100 = 10
number of moles of NaOH = 0.3 *25 = 7.5
pH = pKa + log(salt/acid)
pH = -log(4.8*10^-4) + log(7.5/10)
pH = 3.194
number of moles of HNO2 = 0.1 *100 = 10
number of moles of NaOH = 0.3 *33.333 = 9.9999
pH = 7 + 1/2( pKa + log C)
pH = 7 + 1/2( -log(4.8*10^-4) + log((0.3*33.333)/(100+33.333)))
pH = 8.097
number of moles of HNO2 = 0.1 *100 = 10
number of moles of NaOH = 0.3 *50 = 15
number moles of base remainig = 15 - 10 = 5
pOH = -log(((100-50)/(100+50))*0.3)
pOH = 1
pH = 14 - 1 = 13

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