Question

What is the pH of the resulting solution?

20.0 mL of pure HC2H3O2 is diluted to 1.30 L with water. The density of pure HC2H3O2 is 1.05 g/mL.

Answer 1

See Below

Explanation:

First we are gonna get mL of Acetic Acid into grams. Then get it into moles. Then get it into molarity. Then use the `K_A` for acetic acid to determine `[H^+]`, then pH.

20.0mL of `CH_3COOH xx 1.05g/"mL"` =
21g `CH_3COOH`
21g `CH_3COOH xx ("1mole"/"60.5g") = 0.347 "moles" CH_3COOH`

`(0.347 "moles" CH_3COOH)/(1.3 "liters")` = 0.267M `CH_3COOH`

`CH_3COOH K_A = 1.8xx10^-5`

`CH_3COOH = CH_3COO^- + H^+`

Using an ICE table (and assuming x << 0.267M)

`K_A = 1.85xx10^-5 = x^2/0.267`

x = 0.0022M = `[H^+]`
`pH = -log[H^+]`
`pH = -log(0.0022M)`
`pH = 2.66`