Okey so [H+] = 9.3×10¯5 is given. From given data we can derive the following
Equation for pH is;
pH = - log [H+]
= - log (9.3×10¯5)
= - log 9.3 + log 10¯5
= - (0.9684 + (-5)
= - (-4.0316)
= 4.0316
The pH = 4.0316
(b.)
By using the equation ;
14 = pH + pOH
pOH = 14 - 4.0316 (calculated from above)
pOH = 9.9684
Since pOH = 9.9684
pOH = - log [OH-]
9.9684 = - log [OH-]
[OH-] = antilog (-9.9684)
[OH-] = 1.075 ×10¯(10)