Question

14. 35.00 mL of a 0.30 M HCl solution is titrated with 0.35 M NaOH. What is the pll of the solution after 28.00 mL of the NaOH has been added? (show your calculation steps) - 15. How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M hypochlorous acid to the equivalence point? The Ka of hypochlorous acid is 3.0 x 10-8. (show your calculation steps)

Answer 1

HCl being a strong acid dissociates almost completely into ions. That's the reason why we take no of moles of Hydrogen ions to be equal to Number of excess moles of HCl.

- 14 Hie + NaOH H₂O + Nase - - No of moles = Molarityx Volume (In Litery) Also 12= 1000 ml 3. No of males of Hill= 0.30 malle x 35.00 mL - = 0.30 mallL x 35.00 l = 0.0105 mol 1000 - No of males of NaOH = 0.35 mal x 28.00 h = 0.0098 mal - L 1000 - In the Balanced oxha 1 mole of HCl react with 1 male of NaOH. As no. of males of NaOH are less than no. of moles of Ha so, only 0.0098 mol of Hol will be neutralized by NaOH o Excess moles of Hola 0.0/05 mol 0.0098 mol = 0.0007 modo o Conc. of H+ ion in solution after heutralization is - given by : [H²) = Excess moles = 0.0007 mol Total volume 35.00+28.00 me = 0.0007 x_tooo M = 0.0114 M 63.00 PHz - Dog [H +) = -log(0.01111) = 1.95

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