3 of the 5 parts are done - really need help with the last two!! Help to get the
ID: 870978 • Letter: 3
Question
3 of the 5 parts are done - really need help with the last two!! Help to get these two parts corrected would be greatly appreciated!!
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HCIO(aq) with 0.200 M KOH(aq). The ionization constant for HCIO can be found here. before addition of any KOH after addition of 25.0 mL of KOH(c) after addition of 30.0 mL of KOH after addition of 50.0 mL of KOH after addition of 60.0 mL of KOH With equal concentrations of monoprotic titrant and analyte, the equivalence point would occur when the added volumes are equal (50.0 mL of KOH added). So at point (b), we are 25.0/50.0 = 50% of the way to the equivalence point. If 50% of the add has reacted, then 50% remains, and 50% of the conjugate base has been formed so [CIO-]/[HCIO] = 50/50 = 1. Find the pH using the Henderson-Hasselbalch equation. pH=pKa+log[CLO-]/HCLO]Explanation / Answer
B) after addition of 25 ml of KOH
at mid point pH = pka . In the data you have not mentioned PKa of HClO .
I have taken pKa = 7.1 = pH. you please claculate pKa from the data and take it as pH.
concentration of KOH = 0.2*25/75 = 0.067 M
concentration of HClO = 0.2*50/75 = 0.133 M
net concentration of HClO = 0.133-0.67 = 0.066 M
pH = pKa+log [base/acid]
= 7.1 + log (0.067/0.066)
= 7.1
at mid point pH = pKa
C) after addition of 30 ml of KOH
concentration of KOH = 0.2*30/80 = 0.075 M
concentration of HClO = 0.2*50/80 = 0.125 M
net concentration of HClO = 0.125-0.075 = 0.05 M
pH = pKa+log [base/acid]
= 7.1 + log (0.075/0.05)
= 7.28
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