Answer:
The net ionic equation for the given reaction :

Explanation:
...[1]
..[2]
...[3]

Replacing
, NaI and
in [1] by usig [2] [3] and [4]

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

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Answer:

Explanation:
1. Calculate the initial moles of acid and base

2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)