You need the molarity of HCl. You can take some liberties with the question in as much as it says completely neutralized. You can assume that you have 1 mL of 0.1M NaOH in excess which will be diluted to 0.10M x (1/21) = 0.00476
Then pOH = -log(OH^-), find pH from
pH + pOH = pKw = 14.0
What is the ph of the final solution when 10 mL of HCl is completely neutralized with 11mL of 0.1M Na(OH)?
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