Question

The ph of a solution prepared by the addition of 10. Ml of 0. 002 m koh(aq) to 10. Ml of distilled water is closest to.

Answer 1

This problem is asking for the for pH when a solution of potassium hydroxide is diluted to the double of its initial volume. Thus, the answer turns out to be 11.0.

pH calculations

In chemistry, pH calculations are performed based on the following equation relating the pH and the concentration of hydrogen ions in the solution:

$pH=-log([H^+])$

However, since this problem involves a strong base, KOH as potassium hydroxide, one must calculate the pOH instead, and then convert it to pH with the following equations:

$pOH=-log([OH^-])\\\\pH=14-pOH$

In addition, due to the fact it is diluted from 10. mL to 20. mL (with the 10. mL of added distilled water), one needs to use the dilution equation in order to calculate the correct concentration of KOH:

$M_1V_1=M_2V_2\\\\M_2=\frac{M_1V_1}{V_2} \\\\M_2=\frac{0.002M*10.mL}{20.mL}\\ \\M_2=0.001M$

Hence, the pOH turns out to be:

$pOH=-log(0.001)=3.0$

And thereby the pH:

$pH=14-3.0\\\\pH=11.0$

Learn more about pH calculations: brainly.com/question/1195974