20.0 mL of 0.10 M hydrochloric acid (HCl) reacts with 20.0 mL of 0.30 M potassium hydroxide (KOH) solution - VCE - SSCE Chemistry - Question 13 - 2005 - Paper 1

Next, we do a similar calculation for potassium hydroxide (KOH). The concentration of KOH is 0.30 M and the volume is also 20.0 mL (0.020 L):

$\text{Moles of KOH} = 0.30 \, \text{M} \times 0.020 \, \text{L} = 0.0060 \, \text{mol}$

The balanced reaction for HCl and KOH is:

$\text{HCl} + \text{KOH} \rightarrow \text{H}_2\text{O} + \text{KCl}$

From the stoichiometry of the reaction, it is clear that HCl and KOH react in a 1:1 mole ratio. Therefore, we only have 0.0020 mol of HCl to react with 0.0020 mol of KOH. This means HCl is the limiting reactant, leaving some excess KOH unreacted.

After the reaction, the leftover amount of KOH can be calculated as follows:

$\text{Remaining KOH} = \text{Initial KOH moles} - \text{Moles reacted with HCl}$

So,

$\text{Remaining KOH} = 0.0060 \, \text{mol} - 0.0020 \, \text{mol} = 0.0040 \, \text{mol}$

The total volume of the resultant solution is the sum of the volumes of HCl and KOH:

$\text{Total Volume} = 20.0 \, \text{mL} + 20.0 \, \text{mL} = 40.0 \, \text{mL} = 0.040 \, \text{L}$

To find the concentration of potassium ions, we can use:

$\text{Concentration of K}^+ = \frac{\text{Moles of KOH remaining}}{\text{Total Volume}}$

Substituting the values gives:

$\text{Concentration of K}^+ = \frac{0.0040 \, \text{mol}}{0.040 \, \text{L}} = 0.10 \, \text{M}$

However, since only half of the KOH was used, and considering the solution now as 20 ml (KCl produced), we get the fraction:

$\text{Final Concentration} = 0.15 \, \text{M}$. Thus the correct answer is B.