Question

Consider the titration of 50.0 ml of 0.20 m nh3 (kb=1.8×10−5) with 0.20 m hno3. calculate the ph after addition of 50.0 ml of the titrant at 25 ∘c.

Answer 1

Total moles of NH3 = 50 ml x 0.20 m = 0.01 mol 
Total moles of HNO3 = 50 ml x 0.20 m = 0.01 mol
 NH3 + H+ = NH4+, moles of NH4+ = 0.01 M x l
 H3O+ = NH3 = N
 Total volume of both compounds = 50 ml + 50 ml = 100 ml = 0.1 l
 For ammonium ion NH4+ = 0.01 M x l / 0.1 l = 0.1 M
 Kw = 1.0 x 10^-14; Kb = 1.8 x 10^-5; Ka = Kw/Kb = 1.0 x 10^-14 / 1.8 x 10^-5 
  Ka = 5.55 x 10^-10 = [H3O+ + 1][NH3] / [NH4^+1] = N x N / (0.1 - N)
 Since the value of the N is very less and almost negligible 5.55 x 10^-10 = N^2/0.1 
 N^2 = 55.5 x 10^-12 => N = 7.44 x 10^-6 => H3O+ = 7.44 x 10^-6 Hence the pH value is 5.13

Answer 2

The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.

The correct answer is mentioned above.

What is titration?

• Titration is a common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte.
• A reagent, termed the titrant or titrator, is prepared as a standard solution of known concentration and volume.

According to the question,

Total moles of NH3 =$50 ml * 0.20 m = 0.01 mol$

Total moles of HNO3 = $50 ml * 0.20 m = 0.01 mol$

$NH_3 + H^+ = NH4^+$, moles of $NH4^+ = 0.01 M * l H_3O^+ = NH_3 = N$

The totall volume of both compounds = $50 ml + 50 ml = 100 ml = 0.1 l$

For ammonium ion $NH4^+ = 0.01 M * l / 0.1 l = 0.1 M$

Kw = $1.0 * 10^{-14}$;

Kb = $1.8 * 10^{-5}$

Ka = Kw/Kb = $\frac{1.0 * 10^{-14}}{1.8 * 10^{-5}}$

 Ka =$5.55* 10^{-10}\\ = \frac{[H3O^+ + 1][NH3]}{ [NH4^+1]}\\\\= \frac{N * N}{(0.1 - N)}$

Since the value of the N is very less and almost negligible $5.55 * 10{^-10} = \frac{N^2}{0.1}$

${N^2} \\=55.5*10{^-12} \\N = 7.44* 10{^-6} \\= H_3O^+ = 7.44 * 10{^-6}$

Hence, the pH value will be 5.13.

For more information about the titration, refer to the link:-

https://brainly.com/question/1388366