Question

What is the concentration of a $ KOH\left( {aq} \right) $ solution if $ 12.8 \cdot mL $ of this solution is required to react with $ 25.0 \cdot mL $ of $ .110 \cdot mol/L $ ?

Answer 1

Hint: The potassium hydroxide and sulphuric acid reacts to form potassium sulphate and water. This is a neutralisation reaction which gives salt and water as products. Potassium sulphate is the salt produced here. To solve the given problem we need to find the correct chemical equation for this reaction.

Complete answer:
The proper chemical equation for the reaction between the potassium hydroxide and sulphuric acid is given as - $ 2KOH(aq) + {H_2}S{O_4} \to {K_2}S{O_4}(aq) + 2{H_2}O(l) $ .
Now, based on the given reaction, for calculations, we assume stoichiometric quantities of potassium hydroxide and sulphuric acid.
Now, the moles of sulphuric acid can be calculated as product of volume of sulphuric acid and its concentration, i.e. $ 25.0 \times {10^{ - 3}}L \times 0.110 \cdot mol \cdot {L^{ - 1}}\; = \;2.75 \cdot mol $
Now, as given in the equation, the stoichiometric ratio is $ KOH:{H_2}S{O_4}::2:1 $ , which means that the two mol of potassium hydroxide requires one mol of sulphuric acid. Thus, the moles of potassium hydroxide required for $ \;2.75 \cdot mol $ of sulphuric acid will be $ 2 \times \;2.75 \cdot mol = 5.5 \cdot mol $
Now we have – moles of potassium hydroxide required $ = 5.5 \cdot mol $
And, volume of potassium hydroxide used $ = 12.8 \cdot ml $
The concentration of $ KOH\left( {aq} \right) $ solution will be - $ \dfrac{{mol(KOH)}}{{vol(KOH)}} = \dfrac{{5.5 \cdot mol}}{{12.8 \times {{10}^{ - 3}}L}} = 429.69 \cdot mol \cdot {L^{ - 1}} $
Hence, the concentration of the $ KOH\left( {aq} \right) $ solution is $ 429.69 \cdot mol \cdot {L^{ - 1}} $ .

Note:
The conversion should be done carefully. To find the concentration, first we need to calculate the number of moles of substance and its volume and divide the number of moles by volume. The volume should be taken in litres. The units should be quoted carefully and correctly.