mol Ba(OH)2 = M x L = ??.
(OH^-) from Ba(OH)2 is twice that.
Add to the following.
mol NaOH = M x L = ?
(OH^-) = total mole OH^-/total L
Then pOH = -log(OH^-)
Substitute pOH in the following and solve for pH.
pH + pOH = pKw = 14
Calculate [OH-] and pH for the following strong base solution;
a solution formed by mixing 10.0 mL of 0.011 M Ba(OH)2 with 28.0 mL of 7.4 10-3 M NaOH
working and answer would be appreciated
1 answer
1 answer