Question

1. Titrations are generally both more accurate and more precise the smaller the concentration of titrant you use. A 6.0 M NaOH stock solution is provided by the stock room. (a) What volumes of the stock NaOH solution and DI water would you need to prepare 500 mL of 0.10 M NaOH solution? (b) If you measured your diluted NaOH solution using a pH meter, what pH should it read?

Answer 1

8.3 mL of NaOH Stock, 492 mL of DI Water. pH = 13.

Step-by-step explanation:

Use the M1V1 = M2V2 formula, where m is molarity and v is volume. This can be done in mL or L, it will cancel out.

(6.0)V1 = (.10)(500), solving for V1 you get 8.3 mL of 6.0M NaOH stock solution. The remaining 500-8.3= approx 492 mL should be DI water.

The pH = -log[H+] but in this case we have OH-, so we will use pH + pOH =14. And rearrange to solve for pH = 14 + pOH = 14 + log[OH-]

Then solve

pH = 14 + log(0.10 M0 = 14 - 1 = 13