32 Enough water is added to the buffer in Question 30 to mak

32. Enough water is added to the buffer in Question 30 to make the total volume 5.00 L (a) Calculate the pH of the buffer. (b) Calculate the pH of the buffer after adding 0.0250 mol of HCI to 0.376 L of the buffer.

Solution

32. initial pH of buffer in 30.

pH = pKa + log(base/acid)

     = 4.34 + log[(39.456 mmol/44.693 mmoles)]

     = 4.28

when it is diluted to volume = 5 L

initial volume of buffer = 239 + 137 = 376 ml

final [K2C4H4O6] = (0.288 M x 137 ml)/5000 ml = 0.0079 M

final [KHC4H4O6] = (0.187 M x 239 ml)/5000 ml = 0.0089 M

final pH

pH = 4.34 + log(0.0079/0.0089)

     = 4.28

(b) after HCl = 0.025 mol added

new [K2C4H4O6] = 0.105 M x 0.376 L - 0.025  = 0.0145 mol

final [KHC4H4O6] = 0.120 M x 0.376 L + 0.025 = 0.0701 mol

final pH

pH = 4.34 + log(0.0145/0.0701)

     = 3.65

(c) after KOH = 0.025 mol added

new [K2C4H4O6] = 0.105 M x 0.376 L + 0.025  = 0.0701 mol

final [KHC4H4O6] = 0.120 M x 0.376 L - 0.025 = 0.0145 mol

final pH

pH = 4.34 + log(0.0701/0.0145)

     = 5.02

(e) dilution does not change the pH of the solution as ratio of conjugate base/acid remained the same.