Calculate the pH of the final solution after 100.0 mL of 0.20 MHCl (aq) is titra

Question

Calculate the pH of the final solution after 100.0 mL of 0.20 MHCl(aq) is titrated with 90.0 mL of 0.10 MBa(OH)2. NOTE:Ba(OH)2 is a strong base. ALSONOTE: The system has not yet reached the equivalencepoint.

Explanation / Answer

2HCl(aq) + Ba(OH)2(aq) --> 2H2O(l) + BaCl2(aq) (.1L)(.20M) = .02 moles HCl(aq) (.090L)(.10M) = .009 moles BaCl2      2HCl(aq) + Ba(OH)2(aq) --> 2H2O(l) +BaCl2(aq) I     .02 mol     .009mol             -            0      -.009 mol -.009mol           -           + .009 mol F    .011 mol    0                     -             .009 mol [H+] = .011 mol / .190 L = .05789 M pH = -log[.05789] pH = 1.24 Hope this helps.

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