1. You have a 1L solution of 0.1 M phosphate buffer at pH 2.0. What is the pH of this solution after you add 5 grams of NaOH? Assume there is no volume change and phosphate buffer has the following pKa values: 2.15, 6.82 and 12.38.
The molar ratio of the initial buffer components is calculated:
n H2PO4- / n H3PO4 = 10 ^ (pH - pKa) = 10 ^ (2 - 2.15) = 0.7
It has:
i) n H2PO4- - 0.7 * n H3PO4 = 0
ii) n H2PO4- + n H3PO4 = M * V = 0.1 M * 1 L = 0.1 mol
System of equations between i and ii is applied and you have:
n H2PO4- = 0.041 mol
n H3PO4 = 0.059 mol
The added moles of NaOH are calculated:
n NaOH = g / MM = 5/40 = 0.125 mol
NaOH reacts with all H3PO4 and forms H2PO4-, then reacts with H2PO4- and forms HPO4-2, where it creates the buffer with pKa 6.82, the moles of each remaining species are calculated:
n NaOH = n HPO4-2 = 0.125 - 0.059 = 0.066 mol
n H2PO4- = 0.1 - 0.066 = 0.034 mol
The new pH is calculated:
pH = pKa + log (n HPO4-2 / n H2PO4-) = 6.82 + log (0.066 / 0.034) = 7.11
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