Answer:
The pH is [tex]pH = 9.4[/tex]
Explanation:
From the question we are told that
The volume of [tex]NH_3[/tex] is [tex]V_N = 75mL = 75 *10^{-3} L[/tex]
The concentration of [tex]NH_3[/tex] is [tex]C_N = 0.200M[/tex]
The concentration of [tex]HNO_3[/tex] is [tex]C_H = 0.500 M[/tex]
The volume of [tex]HNO_3[/tex] added is [tex]V_H = 13mL = 13 *10^{-3 } L[/tex]
The base dissociation constant is [tex]K_b = 1.8*10^{-5}[/tex]
The number of moles of [tex]HNO_3[/tex] that was titrated can be mathematically represented as
[tex]n__{H}} = C_H * V_H[/tex]
substituting values
[tex]n__{H}} = 0.500* 13*10^{-3}[/tex]
[tex]n__{H}} = 0.0065 \ moles[/tex]
The number of moles of [tex]NH_3[/tex] that was titrated can be mathematically represented as
[tex]n__{N}} = C_N * V_N[/tex]
substituting values
[tex]n__{N}} = 0.200 * 75*10^{-3}[/tex]
[tex]n__{N}} = 0.015 \ mole[/tex]
So from the calculation above the limited reactant is [tex]HNO_3[/tex]
The chemical equation for this reaction is
[tex]NH_3 + HNO_3 ------> NH^{4+} + NO^{3+}[/tex]
From the chemical reaction
1 mole of [tex]HNO_3[/tex] is titrated with 1 mole of[tex]NH_3[/tex] to produce 1 mole of NH^{4+}
So
0.0065 moles of [tex]HNO_3[/tex] is titrated with 0.0065 mole of [tex]NH_3[/tex] to produce 0.0065 mole of [tex]NH^{4+}[/tex]
So
The remaining moles of [tex]NH_3[/tex] after the titration is
[tex]n = n__{N}} - n__{H}}[/tex]
=> [tex]n = 0.015 - 0.0065[/tex]
[tex]n = 0.0085 \ moles[/tex]
Now according to Henderson-Hasselbalch equation the pH of the reaction is mathematically represented as
[tex]pH = pK_a + log [\frac{NH_3}{NH^{4+}} ][/tex]
Where [tex]pK_b[/tex] is mathematically represented as
[tex]pK_a = -log K_a[/tex]
Now [tex]K_a = \frac{K_w}{K_b}[/tex]
Where [tex]K_w[/tex] is the ionization constant of [tex]NH_3[/tex] with value [tex]K_w = 1.0*10^{-14}[/tex]
Hence [tex]K_a = \frac{1.0*10^{-14}}{1.8 *10^{-5}}[/tex]
[tex]K_a = 5.556 * 10^{-10}[/tex]
Substituting this into the equation
[tex]pH = -log K_a + log [\frac{NH_3}{NH^{4+}} ][/tex]
[tex]pH = log [\frac{\frac{NH_3}{NH^{4+}} }{K_a} ][/tex]
substituting values
[tex]pH = log [\frac{\frac{0.0085}{0.0065} }{5.556*10^{-10}} ][/tex]
[tex]pH = 9.4[/tex]
