Question

ASAP please

Consider the titration of 50.0 mL of 0.200 M hypochlorous acid HCIO (Ka = 3.5 x 10-8) with 0.250 M NaOH. 5- How many milliliters of NaOH are required to reach the equivalence point? a) b) Calculate the pH before titration c) Calculate the pH after adding 40.0 mL. of NaOH d) Calculate the pOH after adding 20.0 ml, of NaOH

Answer 1

50 ml 0.200 (M) Hclo = 50 times .200/1000 = 0.01 mole when no. of equivalent weight of HCl and NaOH will be same, that is the equivalence point Let, x ml NaOH is needed. So, xml 0.250 (M) = x times 0.250/1000 = (2.5 times 10^-4 times x) mole. so, 2.5 times 10^-4 times x = 0.01 x = 40 so, 40 ml of NaOH is needed. - pH before titration [H^+] = squareroot k_a times c = squareroot 3.5 times 10^-8 times .200 = squareroot 7 times 10^-5 = 8.37 times 10^-5 pH = -log [H^+] = -log [8.37 times 10^-5] = 4.08 - when 40 ml NaOH in added then salt hydrolysis is taken place so, pH = 1 /2 p^k w + 1/2 p ka + 1/2 log c pH = 7 + 3.73 + 1/2 log (0.200) pH = 10.73 - 0.3495 pH = 10.38 Ans 20 ml 0.250 (M) NaOH = 20 times .250/1000 = 0.005 mole so, amount of salt = 0.005 mole. amount of remaining acid = 0.005 mole \r\n this is