A solution of HCL (.1053 M) is added to a solution of NaOH (50.0 mL) It takes 28.31 mL if the acid to neutralize the base (naOH).

You don't have a question. I assume you want to know the M of the NaOH.

mLacid x M acid = mLbase x M base.

Sorry I need to neutralization reaction written and I am lost I know it goes into h20 and NaCl but not sure of how to balance it

To determine the concentration of the NaOH solution, we can use the equation:

M1V1 = M2V2

Where:
M1 = concentration of HCl solution
V1 = volume of HCl solution used
M2 = concentration of NaOH solution
V2 = volume of NaOH solution used

Now we can substitute the known values into the equation:

(0.1053 M)(V1) = (M2)(28.31 mL)

We know that the volume of the NaOH solution used is 50.0 mL, so we can replace V2 with 50.0 mL:

(0.1053 M)(V1) = (M2)(28.31 mL)
(0.1053 M)(V1) = (M2)(50.0 mL)

Simplifying the equation, we have:

0.1053 V1 = 50.0 M2

To solve this equation, we need to know either the volume of HCl used (V1) or the concentration of NaOH (M2).

Please provide either the volume of HCl used or the concentration of NaOH to proceed with the calculation.