<h3>
Answer:</h3>
9.17 mL
<h3>
Explanation:</h3>
In this question<u> we are given;</u>
- Molarity of NaOH as 0.150 M
- Molarity of the acid, HCl as 0.055 M
- Volume of the acid, HCl as 25 mL
- Equation for the reaction as;
NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)
We are required to determine the volume of the alkali, NaOH used;
<h3>Step 1: Determine the moles of the acid, HCl</h3>
Molarity = Number of moles ÷ Volume
Thus, rearranging the formula;
Moles = Molarity × volume
= 0.055 M × 0.025 L
= 0.001375 moles
<h3>Step 2: Moles of the alkali, NaOH</h3>
- From the reaction; 1 mole of HCl reacts with 1 mole of NaOH
Therefore;
- Moles of HCl = Moles of NaOH
Hence; Moles of NaOH = 0.001375 moles
<h3>Step 3: Determine the volume of the alkali, NaOH</h3>
- From the previous equation;
Molarity = Moles ÷ Volume
- Rearranging the equation;
Volume = Moles ÷ Molarity
Therefore;
Volume of NaOH = 0.001375 moles ÷ 0.150 M
= 0.00917 L, but, 1 L = 1000 mL
= 9.17 mL
Therefore, the volume of NaOH required is 9.17 mL
