Question

A solution of koh required 31.3 ml of 0.118 m hcl to reach neutralization. what mass of koh was in the titrated sample?

Answer 1

Neutralization reaction is a reaction that involves reaction of a base/alkali and an acid forming salts and water as the only products. In this case;
KOH(aq)+ 2HCl(aq) = H2O(l) + 2KCl(aq)
Moles of HCl used =  (31.3/1000)× 0.118 = 0.00369 moles
Using mole ratio of KOH :HCl = 1:2 , the mole of KOH will be;
  0.00369 ÷ 2= 0.001845 Moles
But, 1 mole of KOH = 56 g
Therefore; 0.001845 moles will have ; 0.001845 × 56 = 0.10332g
Hence the mass of KOH titrated was 0.10332g

Answer 2

Answer:- 0.207 g of KOH.

Solution:- The acid-base reaction is known as neutralization reaction. In general, acid and base reacts to neutralize each other to form salt and water.

HCl is a strong acid and KOH is a strong base. At equivalence point the solution will be neutral. The balanced equation for the reaction of HCl with KOH is written as:

$HCl+KOH\rightarrow KCl+H_2O$

From the above reaction, they react in 1:1 mol ratio. From given molarity and volume, the moles of HCl are calculated and with the help of mol ratio the moles of KOH are calculated that could further be multiplied by its molar mass to get the grams.

Molar mass of KOH = 39.098 g + 16 g + 1.008 g = 56.106 g

The calculations are shown below using dimensional analysis:

$31.3mL(\frac{1L}{1000mL})(\frac{0.118molHCl}{1L})(\frac{1molKOH}{1molHCl})(\frac{56.106gKOH}{1molKOH})$

= 0.207 g KOH

So, the mass of KOH in the titrated sample is 0.207 g.