Calculate the PH of and POH of the following solution and comment on them.(a)0.39 moldm3 of H2SO4(b)2.0 moldm3 of NaoH
For NaOH, this is 100% dissociated; therefore, (OH^-) = (NaOH)
OH^- = 2 M
pOH = -log(OH^-)
pH + pOH = pKw = 14.
You know pOH, solve for pH.
The H2SO4 isn't as simple as it lookis. If you are an advanced student,
(0.39+x)/(0.39-x) = k2 for H2SO4.
Solve for x and evaluate 0.39+x then
pH = -log(H^+). If you are not an advanced student I assume the prof wants you to answer
(H^+) = 2*0.39, then
pH = -log(H^+).