Question
.012 mol of solid NaOH is added to a 1 L solution of HCl 0.01 M. What is the pH of the solution?
Answers
DrBob222
mols HCl = M x L = 0.01 x 1 = 0.01
mols NaOH = 0.12
mols NaOH after neutralization = 0.11.
M NaOH = M OH^- = mols/L.
Then pOH = -log(OH^-) and
pH + pOH = pKw = 14.
Given pOH and pKw, solve for pH.
mols NaOH = 0.12
mols NaOH after neutralization = 0.11.
M NaOH = M OH^- = mols/L.
Then pOH = -log(OH^-) and
pH + pOH = pKw = 14.
Given pOH and pKw, solve for pH.