O 4/25/2015 06:00 PM A 67.1/100 4/13/2015 11 8 AM Gradebook Print Calculator Per

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O 4/25/2015 06:00 PM A 67.1/100 4/13/2015 11 8 AM Gradebook Print Calculator Periodic Table Question 12 of 19 Map A General Chemistr sity Science Books presented by Sapling Le Donald McQuarrie. Peter A Rock .Ethan Gallogly Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.230 M HCIO(ag) with 0.230 M KOH(aq). The ionization constant for HCIO can be found here. Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH Number (c) after addition of 40.0 mL of KoH Number (d) after addition of 50.0 m of KOH Number (e) after addition of 60.0 mL of KOH B Hint e Previous Give Up & View Solution Check Answer Next Ext

Explanation / Answer

Answer: Here a] before addition of any KOH means the only 50ml of 0.23 M HclO is present hence the Ph = -log[H+]

Ph = - log [0.23 ] = 0.638 acidic medium

b and all part} Here when we add some base KOH means there is formation of salt and also at the end point the number of moles of acid = number of moles of base

and now we have to use the Henderson's equation

Ph = PKa + log [Salt]/[Acid remains]

Now we adding 25 ml of base means we have number of moles of base in 25 ml = 0.23 * 25 / 1000 = 0.00575 mol

Hence the 0.00575 mol of acid and base react to form 0.00575 mol of salt

And the number of moles of acids in 50 ml is = 0.23 * 50 /1000 = 0.015 mol

Hence the amount of acids unreacted is = 0.015 - 0.00575 = 0.00575 mol

Now put all the values in Hendersons equation we get the required Ph

Ph = PKa + log [salt]/[acid]

The Ka values should be given in question and by using the given value first calculate the PKa = -log[Ka] values

and put the all calculated moles we get the required Ph values

Note: The same procedure is for all the parts .

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