Question

Be sure to answer all parts.

Find the pH and the volume (mL) of 0.407 M HNO3 needed to reach the equivalence point in the titration of 2.65 L of 0.0750 M pyridine (C5H5N, Kb = 1.7 × 10−9).
Volume = mL HNO3
pH =

Answer 1

pH = 3.215

Step-by-step explanation:

From the given information;

Using the equation for the dilution of a stock solution:

Since moles of C5H5N = moles of HNO3

Then:

`M_(C_5H_5N)* V_(C_5H_5N)= M_(HNO_3)* V_(HNO_3)`

`0.0750 M * 2.65 L = 0.407 M * V_(HNO_3)`

`V_(HNO_3)= (0.0750 M * 2.65 L)/(0.407 M)`

`V_(HNO_3)=488.33 \ mL`

The reaction between C5H5NH and H2O is as follows:

`C_5H_5N^+H + H_2O` ⇄
`C_5H_5N + H_3O^+`

`Molarity \ of \ C_5H_5N^+H = (moles)/(addition \ of\ the\ total\ volume)`

`Molarity \ of \ C_5H_5N^+H = (0.0750 \ M * 2.65 \ L)/(2.65 \ L + 0.48833 \ L)`

`Molarity \ of \ C_5H_5N^+H = (0.19875 \ ML)/(3.13833 L )`

`= \ 0.06333\ M`

Now, the next step is to draw out the I.C.E table.

`C_5H_5N^+H + H_2O` ⇄
`C_5H_5N + H_3O^+`

I 0.06333 0 0

C - x x x

E 0.06333 -x x x

`K_a = (10^(-14))/(1.7 * 10^(-19)) \\ \\ K_a = 5.8824 * 10^(-6)`

`K_a = ([C_5H_5N][H_3O^+] )/([C_5H_5N^+H] ) \\ \\ 5.8824 * 10^(-6) = (x^2)/(0.06333 - x)`

Assuming x < 0.06333

`x^2 = 5.8824 * 10^(-6) * 0.06333`

Then

`x^2 = 3.72532392 * 10^(-7) \\ \\ x= \sqrt{3.72532392 * 10^(-7)} \\ \\ x = 6.1035 * 10^(-4) \ M`

`[H_3O^+] = x = 6.1035 * 10^(-4) \ M \\ \\ pH = -log (6.1035 * 10^(-4)) \\ \\ \mathbf{\\ pH = 3.215}`