Question

what is the net ionic equation for this reaction? 2koh(aq) + h₂so₄(aq) → 2h₂o(ℓ) + k₂so₄(aq) a. 2koh + 2h⁺ + so₄²⁻ → 2h₂o(ℓ) + 2k⁺ + so₄²⁻ b. 2koh + h₂so₄ → 2h₂o + k₂so₄ c. oh⁻ + 2h⁺ → 2h₂o(ℓ) d. 2k⁺ + oh⁻ + 2h⁺ + so₄²⁻ → 2h₂o(ℓ) + 2k⁺ + so₄²⁻

Explanation:

Step1: Write the complete ionic equation

The strong - electrolytes $KOH$, $H_2SO_4$, and $K_2SO_4$ dissociate in aqueous solution.
$2K^+(aq)+2OH^-(aq)+2H^+(aq)+SO_4^{2 - }(aq)
ightarrow2H_2O(l)+2K^+(aq)+SO_4^{2 - }(aq)$

Step2: Identify and cancel the spectator ions

The spectator ions are $K^+$ and $SO_4^{2 - }$ as they appear on both sides of the equation.
After canceling, we get $2OH^-(aq)+2H^+(aq)
ightarrow2H_2O(l)$. We can simplify the coefficients by dividing each by 2, resulting in $OH^-(aq)+H^+(aq)
ightarrow H_2O(l)$. But considering the stoichiometry from the original balanced equation before simplifying the complete - ionic coefficients, the correct net - ionic equation is $OH^- + 2H^+
ightarrow2H_2O(l)$

Answer:

C. $OH^- + 2H^+
ightarrow2H_2O(l)$