Question

1) Calculate the pH of a solution that is 1.00 M HNO2 and 1.00 M NaNo2?

2) Calculate the pH after .10 mol NaOh is added to 1.00 L of the solution ?

Answer 1

1)

Concentration of HNO2 = 1.00M

concentration of NaNO2 = 1.00M

Ka of HNO2 = 7.1x10^-4

Ka = 7.1x10^-4

-log(Ka) = -log(7.1x10^-4)

Pka = 3.15

PH = Pka + log[salt]/[acid]

PH = 3.15 + log(1.00/1.00)

PH = 3.15

2)

HNO2 = 1.0L of 1.00M

number of moles of HNO2 = 1.00M x 1.0L = 1.00 moles

NaNO2 = 1.0L of 1.00M

number of moles fo NaNO2 = 1.00M x 1.0L = 1.00 moles

number of moles of NaOH = 0.10 mole

after addition of NaOH

number of moles of HNO2 = 1.00 - 0.100 = 0.900 mole

number of moles of NaNO2 = 1.00 + 0.100 = 1.1 MOLE

PH = 3.15+ log(1.1/0.9)

PH = 3.24