Question

Part A

Calculate the change in pH that results from adding 0.190 mol NaNO2 to 1.00 L of 0.190 M HNO2(aq). (Ka(HNO2)=7.2⋅10−4) Express your answer using two decimal places.

Part B

Calculate the change in pH that results from adding 0.190 NaNO3 to 1.00 L of 0.190 HNO3(aq). Express your answer using two decimal places.

Answer 1

Answer – Part A) we are given, 1.0 L of 0.190 M HNO₂, Ka = 7.2*10^-4

pH before adding the NaNO₂

we know HNO₂ is weak acid –

HNO₂ + H₂O ------> H₃O⁺ + NO₂^-

I 0.190                      0           0

C -x                         +x          +x

E 0.190-x                   +x          +x

Ka = [H₃O⁺] [NO₂^-] / [HNO₂]

7.2*10^-4 = x*x /(0.190-x)

7.2*10^-4 *(0.190-x) = x²

Now we need to set up quadratic equation

1.37*10^-4 - 7.2*10^-4 x = x²

x² + 7.2*10^-4 x – 1.37*10^-4 = 0

a =1 , b = 7.2*10^-4 , c = -1.37*10^-4

Using the quadratic equation

x = -b +/- √b²-4a*c / 2a

Plugging the value in this formula

x = 0.0113 M

so, x = [H₃O⁺] = 0.0113 M

so, pH = -log [H₃O⁺]

           = -log 0.0113 M

           = 1.95

pH after adding the NaNO₂

we know, moles of NaNO₂ = moles of NO₂^- = 0.190 moles

[NO₂^-] = 0.190 moles / 1.0 L = 0.190 M

So, [HNO₂] = [NO₂^-]

So we know, when acid and its conjugate base has same molarity then ration of conjugate base by acid is 1 and pH = pKa

So, pKa = -log Ka

             = - log 7.2*10^-4

             = 3.14

So, pH = 3.14

Chang in pH = 3.14 – 1.95

                     = 1.20

Part B) We are given, [HNO₃] = 0.190 M , volume = 1.0 L

We know HNO₃ is strong acid , so

[HNO₃] =[H₃O⁺] = 0.190 M

So, pH = -log [H₃O⁺]

           = -log 0.190 M

           = 0.72

pH after addition of 0.190 moles of NaNO₃- When we added 0.190 M NaNO₃ then there is no any change in the [H₃O⁺], so pH remains same, so no any change.

So change in pH = 0.00