Question

Calculate the pH of the solution to 10 liters which contained 3.7 g of potassium

hydroxide​

Answer 1

Given, 3.7 grams of KOH are present in the solution having volume 10 litres.

Steps involved:

1) Find the molarity of OH- ions.

2) Find the POH of the solution with the formula POH = -log(base 10) OH- .

3) Find the PH of the solution by subtracting POH from 14. (PH + POH=14) .

Step 1:

=> Molar mass of KOH is 39+16+1 =>56 grams per mole.

So, 3.7 grams of KOH is 3.7/56 => 0.0661 moles (approx.)

Each mole of KOH in aqueous state gives 1 mole of OH-1 ions.

So, 0.0661 moles of OH-1 will be produced in aqueous state.

volume of solution= 10 litres.

So, molarity = (moles of OH-1)/ (litres of solution)

molarity= 0.0661/10 => 0.00661 M (molar)

Step 2:

POH = -log (base 10) OH-1 concentration

POH = -log(base 10) 0.00661 (from calculator log 0.00661=  -2.1798)

POH = -1 * -2.1798 => POH = 2.1798

Step 3:

So, PH + POH =14

PH + 2.1798 =14

PH= 14-2.1798

PH= 11.8202

Hence, the PH is 11.82 approximately.

So, the PH of a solution containing 3.7 grams of KOH in 10L of H2O is 11.82