Sure, let's solve this problem step-by-step.
1. Understand the neutralization reaction:
- The reaction between phosphoric acid (H₃PO₄) and potassium hydroxide (KOH) is usually represented as:
[tex]\[ \text{H}_3\text{PO}_4 + 3\text{KOH} \rightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O} \][/tex]
- This tells us that one mole of phosphoric acid (H₃PO₄) neutralizes three moles of potassium hydroxide (KOH).
2. Calculate the moles of phosphoric acid:
- We are given that the volume of phosphoric acid is 78 mL and its concentration is 2.5 M:
[tex]\[ \text{Moles of H}_3\text{PO}_4 = \text{Volume (L)} \times \text{Concentration (M)} \][/tex]
- Convert the volume from mL to L:
[tex]\[ 78 \text{ mL} = 0.078 \text{ L} \][/tex]
[tex]\[ \text{Moles of H}_3\text{PO}_4 = 0.078 \text{ L} \times 2.5 \text{ M} = 0.195 \text{ moles} \][/tex]
3. Determine the moles of KOH required:
- From the balanced reaction, 1 mole of H₃PO₄ reacts with 3 moles of KOH:
[tex]\[ \text{Moles of KOH required} = 3 \times \text{Moles of H}_3\text{PO}_4 \][/tex]
[tex]\[ \text{Moles of KOH required} = 3 \times 0.195 \text{ moles} = 0.585 \text{ moles} \][/tex]
4. Calculate the concentration of the KOH solution:
- We know the volume of the KOH solution is 500 mL:
[tex]\[ \text{Volume of KOH solution in L} = 500 \text{ mL} = 0.5 \text{ L} \][/tex]
[tex]\[ \text{Concentration of KOH} = \frac{\text{Moles of KOH}}{\text{Volume (L)}} \][/tex]
[tex]\[ \text{Concentration of KOH} = \frac{0.585 \text{ moles}}{0.5 \text{ L}} = 1.17 \text{ M} \][/tex]
So, the concentration of the potassium hydroxide (KOH) solution is 1.17 M.
