35.00 mL of a 0.30 M HCl solution is titrated with 0.35 M NaOH. What is the pH of the solution after 40.00 mL of the NaOH has been added

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The correct answer is E.Explanation:The balanced chemical equation for the reaction is:HCl + NaOH → NaCl + H2OThe moles of HCl initially present are:moles of HCl = (0.30 M) x (0.035 L) = 0.0105 molThe moles of NaOH added are:moles of NaOH = (0.35 M) x (0.040 L) = 0.014 molSince NaOH is in excess, all of the HCl will react and the moles of NaOH remaining will be:moles of NaOH remaining = 0.014 mol - 0.0105 mol = 0.0035 molThe total volume of the solution after the addition of NaOH is:total volume = 35.00 mL + 40.00 mL = 75.00 mLThe concentration of NaOH remaining is:[NaOH] = (0.0035 mol) / (0.075 L) = 0.0467 MThe pOH of the solution is:pOH = -log[OH-] = -log(0.0467) = 1.33The pH of the solution is:pH = 14 - pOH = 14 - 1.33 = 12.67

3500 ML of a 0