Potassium hydroxide (KOH) is a strong base. Dissociation of potassium hydroxide follows the reaction equation given. KOH(s) –H20–> K+(aq) OH-(aq)  a) Write down the reaction equation for a neutralization reaction where potassium hydroxide (KOH) reacts with sulphuric acid (H2SO4) producing potassium sulphate (K2SO4) and water (H2O). Balance if needed! b) Calculate pH for the solution where 0,2 mol of KOH is dissolved into water so that the total volume of the solution is 0,5 L.ÂÂ
Potassium hydroxide (KOH) is a strong base. Dissociation of potassium hydroxide follows the reaction equation given. KOH(s) –H20–> K+(aq) OH-(aq)
ÂÂ
a) Write down the reaction equation for a neutralization reaction where potassium hydroxide (KOH) reacts with sulphuric acid (H2SO4) producing potassium sulphate (K2SO4) and water (H2O). Balance if needed!
b) Calculate pH for the solution where 0,2 mol of KOH is dissolved into water so that the total volume of the solution is 0,5 L.ÂÂ
The Correct Answer and Explanation is :
a) Reaction equation for the neutralization between KOH and H₂SO₄:
The neutralization reaction between potassium hydroxide (KOH) and sulfuric acid (H₂SO₄) forms potassium sulfate (K₂SO₄) and water (H₂O). The balanced reaction is:
[
2 \, \text{KOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{K}_2\text{SO}_4(aq) + 2 \, \text{H}_2\text{O}(l)
]
This balanced equation shows that two moles of KOH react with one mole of H₂SO₄ to form one mole of potassium sulfate and two moles of water.
b) pH Calculation for KOH Solution:
To calculate the pH of a solution where 0.2 mol of KOH is dissolved in 0.5 L of water, follow these steps:
- Determine the concentration of OH⁻ ions:
Since KOH is a strong base, it dissociates completely in water:
[
\text{KOH} \rightarrow \text{K}^+ + \text{OH}^-
]
The concentration of OH⁻ ions will be the same as the concentration of KOH, which is:
[
[\text{OH}^-] = \frac{0.2 \, \text{mol}}{0.5 \, \text{L}} = 0.4 \, \text{M}
]
- Calculate the pOH:
The pOH is calculated using the formula:
[
\text{pOH} = -\log[\text{OH}^-]
]
Substituting the concentration of OH⁻:
[
\text{pOH} = -\log(0.4) \approx 0.398
]
- Calculate the pH:
The relationship between pH and pOH is given by:
[
\text{pH} + \text{pOH} = 14
]
Thus:
[
\text{pH} = 14 – \text{pOH} = 14 – 0.398 = 13.602
]
Final answer:
The pH of the solution is approximately 13.6. This indicates a strongly basic solution, as expected for a strong base like KOH.
Would you like an illustration for this process?