A75.0-ml volume of 0.200 m nh3 (kb=1.8×10−5) is titrated with 0.500 m hno3. calculate the ph after the addition of 19.0 ml of hno3. express your answer numerically.

pH after the addition of 19.0 ml of HNO₃ : 9.018

The pH value of a reaction between strong acid HNO₃ and weak base NH₃ can be estimated from the rest of the reaction product

1. If the remainder of the reaction results obtained the remaining strong acid HNO₃, the pH is sought from the concentration of [H⁺] using the formula

[H⁺] = a. M

a = valence of acid / amount of H⁺ released

M = acid concentration

2. When strong acids and weak bases react, they form salts that are acidic, calculating the pH using the pH hydrolysis formula

where

M = concentration of salt anion

3. If the remainder of the reaction results are a weak base remaining and the salt, the solution will form a base buffer solution and search for pH using the formula base buffer pH

We count the moles of each reactant:

NH₃ mole = 75 ml x 0.2 M = 15 mlmol

mole HNO₃ = 19 ml x 0.5 M = 9.5 mlmol

NH₃ + HNO₃ ---> NH₄NO₃

15        9.5    

9.5      9.5            9.5

5.5       0              9.5

so there are remaining weak bases NH₃ = 15 - 9.5 = 5.5 mmol

Then a buffer solution is formed

[OH⁻] = Kb x [weak base mole] / [salt mole x valence]

the pH of a solution

link

Calculate the pH

link

the pH of a 2.0 M solution of HClO4

link

Keywords : pH, acid, base, HNO₃,NH₃

28528289 29823282 8298232 39858