determine h30+ in M, of 10.00ml of .100M NaOH after adding 10.00ml of indicator solution?

To determine the concentration of $\ce{H3O+}$ (hydronium ions) in the solution after adding the indicator solution, we need to consider the reaction that occurs between $\ce{NaOH}$ and $\ce{H3O+}$.

First, let's write the balanced chemical equation for the reaction between $\ce{NaOH}$ and $\ce{H3O+}$:

$\ce{NaOH + H3O+ -> H2O + Na+}$

From the equation, we can see that one mole of $\ce{NaOH}$ reacts with one mole of $\ce{H3O+}$ to form one mole of $\ce{H2O}$. Therefore, the stoichiometric ratio between $\ce{NaOH}$ and $\ce{H3O+}$ is 1:1.

Now, let's calculate the number of moles of $\ce{NaOH}$ in the 10.00 mL of 0.100 M solution:

$\text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)}$

$\text{Moles of NaOH} = 0.010 \, \text{L} \times 0.100 \, \text{M} = 0.001 \, \text{mol}$

Since the stoichiometric ratio between $\ce{NaOH}$ and $\ce{H3O+}$ is 1:1, the number of moles of $\ce{H3O+}$ will also be 0.001 mol.

Now, let's determine the final volume of the solution after adding the 10.00 mL of indicator solution. Since no other information is given, we will assume that the volumes are additive:

$\text{Final volume of solution} = \text{Volume of NaOH} + \text{Volume of indicator solution}$

$\text{Final volume of solution} = 10.00 \, \text{mL} + 10.00 \, \text{mL} = 20.00 \, \text{mL}$

Since the indicator solution is added to a total volume of 20.00 mL, the concentration of $\ce{H3O+}$ will be:

$\text{Concentration of H3O+} = \frac{\text{Moles of H3O+}}{\text{Volume (L)}}$

$\text{Concentration of H3O+} = \frac{0.001 \, \text{mol}}{0.020 \, \text{L}} = 0.050 \, \text{M}$

Therefore, the concentration of $\ce{H3O+}$ in the solution after adding 10.00 mL of the indicator solution is 0.050 M.