If the PH between the reaction of 50ml of 0.3M of excess NaOH with 10ml of HCl is 12, what is the initial concentration of the acid (a) 0.30M (b) 1.44M (c) 0.44M (d) 1.30M

Given $\text{pH} = 12$.
The relationship between pH and pOH is $\text{pH} + \text{pOH} = 14$.

$$\text{pOH} = 14 - \text{pH} = 14 - 12 = 2$$

The volume of $\text{NaOH}$ is $50 \text{ mL} = 0.050 \text{ L}$.
The volume of $\text{HCl}$ is $10 \text{ mL} = 0.010 \text{ L}$.

$$\text{Total Volume} = 0.050 \text{ L} + 0.010 \text{ L} = 0.060 \text{ L}$$

$$\text{Initial moles of NaOH} = \text{Concentration of NaOH} \times \text{Volume of NaOH}$$ $$\text{Initial moles of NaOH} = 0.3 \text{ M} \times 0.050 \text{ L} = 0.015 \text{ mol}$$

The moles of $\text{NaOH}$ that reacted are the initial moles minus the excess moles.

$$\text{Moles of NaOH reacted} = \text{Initial moles of NaOH} - \text{Moles of excess OH}^-$$ $$\text{Moles of NaOH reacted} = 0.015 \text{ mol} - 0.0006 \text{ mol} = 0.0144 \text{ mol}$$

Since $\text{HCl}$ and $\text{NaOH}$ react in a 1:1 molar ratio ($\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}$), the moles of $\text{HCl}$ initially present are equal to the moles of $\text{NaOH}$ that reacted.

$$\text{Moles of HCl} = 0.0144 \text{ mol}$$

$$\text{Initial concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}}$$ $$\text{Initial concentration of HCl} = \frac{0.0144 \text{ mol}}{0.010 \text{ L}} = 1.44 \text{ M}$$

The initial concentration of the acid is $1.44 \text{ M}$. This matches option (b).

The final answer is $\boxed{\text{b) 1.44M}}$.