Consider the titration of 1.0 L of 1.0 M NH3 with 1.0 M HCl. Which of the following correctly describe(s) how the pH would be calculated at each of the following additions of HCl? I: At 0 L HCl, pH would be calculated based on the concentration and Kb of the weak base II: At 1 L HCl, pH would be calculated based on the concentration and Ka of the conjugate acid III: At 2 L HCl, pH would be calculated based on the concentration of excess acid in solution Group of answer choices I and III Only II Only I I, II, and III II and III

I, II, and III

Explanation:

In the titration of NH₃ with HCl:

NH₃ + HCl → NH₄⁺ + Cl⁻

Where NH₃ is the weak base and NH₄⁺ is the conjugate acid.

I: At 0 L HCl, pH would be calculated based on the concentration and Kb of the weak base: At 0L of HCl, you will have just NH₃ in solution. That means you would calculate the pH just from the concentration of the weak base using Kb. That means I is true.

II: At 1 L HCl, pH would be calculated based on the concentration and Ka of the conjugate acid: When you add 1L of HCl, you will have in solution just NH₄⁺, the conjugate acid. That means you would calculate the pH of the solution just with the Ka of the conjugate acid and its concentration. II is true.

III: At 2 L HCl, pH would be calculated based on the concentration of excess acid in solution: At 2L of HCl solution, you have HCl in excess in the solution. As HCl is a strong acid, the pH would be affected in the big way by this concentration in excess. III is true.

i believe that it is a exoplanet

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml