Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added…
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Question “Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added…”
Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added to (a) 33.0 mL of 0.230 M NaOH(aq). pH = (b) 13.0 mL of 0.330 M NaOH(aq). pH =
Answer
Add 23.0mL 0.230M HCI(aq), to 33.0mL 0.230M NaOH(aq) and calculate the pH of the solution. Number pH-12.61 (b) 13.0 mg of 0.330 MNOH(aq). Number pH- moles HCl = 0.230 M x (0.23 L) =0.00529 mol Moles of NaOH (0.230M) x (0.33 L) 0.00759 Mol Here, NaOH is excessive. The solution is acidic excess NaOH = [OH] = (0.00759 -0.00529)/ (0.056L) = 0.0.0410 M pH 14-pOH 12.61 Moles of HCl = (0.230M) x (0.0.023 L), = 0.00529 moles of NaOH (0.033 L) 0.00429 moles Here, HCl is excessive. Acidic excess HCIH+ = (0.00529-0.00429)/(0.036 L) = 0.013 L) 0.00429 mol Here; HCl is in excess
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