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Consider 2.50 L of a buffer solution made from NaOH and 3.50 M H3PO4 at pH...

Consider 2.50 L of a buffer solution made from NaOH and 3.50 M H3PO4 at pH 2.23 that has a total concentration of phosphate containing species of 0.188 M.

How many mL of 2.00 M HCl must be added to the buffer solution to change its pH by 1.25 units?

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Answer #1

initiall

pH = pKa + log(H2PO4- / H3PO4)

2.23 = 2.12 + log( H2PO4- / H3PO4)

H2PO4- / H3PO4 = 10^(2.23-2.12) = 1.288

H2PO4- + H3PO4 =0.188*2.5 = 0.47

H2PO4- = 1.288*H3PO4

1.288*H3PO4 - + H3PO4 = 0.47

H3PO4 = 0.47 / (1.288+1) = 0.20541

H2PO4- = 1.288*0.20541 = 0.26456

now...

assume "x" is the mol amount of HCl

pHnew:

pH = pKa + log(H2PO4- / H3PO4)

2.23-1.25 = 2.12 + log((0.26456-x)/(0.20541+x))

10^(2.23-1.25 -2.12 ) = (0.26456-x)/(0.20541+x)

0.07244 * ( 0.20541) + 0.07244 x = 0.26456-x

(0.07244 * ( 0.20541) - 0.26456) = (-0.07244 -1)*x

x = (0.07244 * ( 0.20541) - 0.26456) / (-0.07244 -1)

x = 0.2328

mol of HCl --> 0.2328

V = mol/V = (0.2328/2) = 0.1164 L= 116.4 mL required

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