A 21.18 mL of 0.250 M NaOH is titrated with a H2SO4 solution. The initial

volume of H2SO4 was 13.28 and the final volume of H2SO4 was 28.29 mL when
the solution turned very slightly pink. What is the concentration of NaOH?

molesacid=2molesbase

Va*Ma=2*Vb*Mb
(28.29-13.28)mL*Ma=2*21.18mL*.250M

solve for Ma

I don't understand what you just wrote

To find the concentration of NaOH, we can use the concept of stoichiometry and the equation of the reaction between NaOH and H2SO4, which is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

Here's how we can calculate the concentration of NaOH:

1. Calculate the moles of H2SO4 used:
Since the initial volume of H2SO4 was 13.28 mL and the final volume was 28.29 mL, the volume of H2SO4 used in the titration can be calculated as:
Volume used = Final volume - Initial volume = 28.29 mL - 13.28 mL = 15.01 mL

Convert the volume used to liters:
Volume used = 15.01 mL * (1 L / 1000 mL) = 0.01501 L

2. Using the balanced equation, determine the mole ratio between NaOH and H2SO4. From the equation, we know that 2 moles of NaOH react with 1 mole of H2SO4.

3. Calculate the moles of NaOH:
Since the mole ratio is 2:1, the number of moles of NaOH used will be half of the number of moles of H2SO4 used:
Moles of NaOH = (1/2) * Moles of H2SO4

To find the moles of H2SO4, we need to use the Molarity(M) and volume(V) relationship:
Moles of H2SO4 = Molarity of H2SO4 * Volume of H2SO4 (in liters)
Moles of H2SO4 = 0.250 M * 0.01501 L

4. Calculate the moles of NaOH:
Moles of NaOH = (1/2) * 0.250 M * 0.01501 L

5. Calculate the concentration of NaOH:
Concentration of NaOH = Moles of NaOH / Volume of NaOH (in liters)
Volume of NaOH is given as 21.18 mL, so we need to convert it to liters:
Volume of NaOH = 21.18 mL * (1 L / 1000 mL)

Concentration of NaOH = ((1/2) * 0.250 M * 0.01501 L) / (21.18 mL * (1 L / 1000 mL))

Now, let's substitute the values into the equation and calculate the concentration of NaOH:

Concentration of NaOH = ((1/2) * 0.250 * 0.01501) / (21.18 * 10^-3)

Concentration of NaOH ≈ 0.1867 M

Therefore, the concentration of NaOH in the given solution is approximately 0.1867 M.