500. ml of a solution containing 1. 5 m nh4cl(aq) is mixed with 500. ml of a solution containing 0. 50m of naoh(aq). what is the

Question

2 years ago

6

ph of the final solution? (kb(nh3) = 1. 8 x 10-5)

1 answer:

Sladkaya [172] · Answer 1 · 2023-04-09T14:08:47+03:00

The pH of the solution is determined by the hydrogen ion concentration. The pH of the solution with 1.5 M NH4Cl and 0.50 M NaOH is 8.77.

The potential of the hydrogen ion in the given solution is called pH which determines the acidity and basicity of the solution.

Given

The molar concentration of 500 mL NH4Cl = 1.5 M

The molar concentration of 500 mL NaOH= 0.50 M

Kb (NH3) = 1.8 × 10 -5

First pOH is calculated as,

pOH = pKB + log [NH4Cl][NaOH]

= - log 1.8 × 10-5 + log [1.5][0.50]

= - log 1.8 × 10-5 + log [3]

= 4.744 + 0.477

= 5.221

Now, pH is calculated from pOH as,

pH + pOH = 14

pH = 14 – pOH

= 14 – 5.221

= 8.77

Thus we found that, 8.77 is the pH of the solution with 1.5 M NH4Cl and 0.50 M NaOH.

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