When writing a lab report where do your data tables belong? conclusion

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We should add 10 mL of the strong base.

Explanation:

Assuming the acid is monoprotic, then the number of H⁺ moles in 100 mL of the solution with pH=5 would be:

5 = -log[H⁺]-5 = log[H⁺] = [H⁺]1x10⁻⁵ M * 0.100 L = 1x10⁻⁶ moles H⁺

In order to make the solution neutral, we should add the same amount of OH⁻ moles (1x10⁻⁶).

To calculate [OH⁻] of the strong base, we can use pOH:

pH + pOH = 1410 + pOH = 14pOH = 4

Once again, assuming the strong base is monobasic:

pOH = -log[OH⁻]4 = -log[OH⁻]10⁻⁴ = [OH⁻]

Finally we calculate the required volume:

1x10⁻⁶ moles OH⁻ ÷ 1x10⁻⁴M = 0.010 L 0.010 L ⇒ 0.010 * 1000 = 10 mL