Question

how to do calculation for and not plug in -log() in calculator

[H3O+] = 3.2 * 10 ^ -4
i wan to find pH and OH-

Answer 1

well, 3.2^2 is about 10, so 3.2 is approx. 10^(1/2)

so log(3.2) is approximately 0.5

now pH
=-log([H3O+] )
= -log(3.2 * 10 ^ -4)
= -log(3.2) - log(10^-4) because log(a*b)= log a + log b
= -0.5 + log(10^4) because log(10^-4) = -log(10^4)
= -0.5 + 4 = 3.5

at 25 deg Celsius, pOH = 14 - pH = 10.5

Answer 2

ty