How to calculate how much base to add to a solution get a desired pH

Question

If a 1.0L solution has an initial pH of 6.70 how much KOH do i need to add to get a desired pH of 6.80?

The solution will react 1:1 with the KOH: H⁺ + OH^- --> H₂O

Answer 1

The initial pOH is 14 - 6.70 = 7.30

The final pOH is 14 - 6.80 = 7.20

7.30= -log(initial concentration of OH)

-7.30 = log(initial concentration of OH)

The initial concentration of OH^- = 10^-7.3M

Similarly, the final concentration of KOH = 10^-7.2M

Due to the 1:1 ratio between KOH and OH^-, the concentration of KOH is 10^-7.3M

The final concentration of KOH is 10^-7.2M

Because there’s 1L of solution, there is initially 10^-7.3 moles of KOH and at the end, 10^-7.2 moles of KOH

The amount added would therefore be 10^-7.2 - 10^-7.3 moles of KOH