Suppose that you are given a 0.700 M acetic acid solution and a 0.500 M sodium a

Question

Suppose that you are given a 0.700 M acetic acid solution and a 0.500 M sodium acetate solution. Determine the volumes of each solution required to prepare 160mL of a buffer with pH 4.65.

Explanation / Answer

This is a buffer, or 'common ion', question so we need to use theHenderson-Hasselbalch equation, which relates the ratio of acid tobase with pH and pKa:     pH = pKa + log([A-]/[HA]) The pKa of acetic acid is about 4.75 (Source:http://www.chembuddy.com/?left=BATE&right=dissociation_constants):    4.65 = 4.75 +log([A-]/[HA])             -0.1 = log([A-]/[HA])     [A-]/[HA] = 10-0.1 Note that these concentrations are the finalconcentrations of acid and base (i.e. you cannot simply plug in thegiven concentrations - it gets you nowhere!). Recall that concentration = moles / volume. This means we nowhave:     10-0.1 = (n(A-)/0.160 L) / (n(HA)/0.160 L)              = n(A-) / n(HA) Where 'n' is the number of moles. Since the number of moles of eachcomponent in the final solution are the same as the number of molestaken from each stock solution, we can write:     10-0.1 = ([A-]0 * v(A-)) /([HA]0 * v(HA))     10-0.1 = (0.5 * v(A-)) / (0.7 *v(HA)) (This is because we can rearrange c = n/v --> n = c*v) The subscript '0' indicates the initial (stock)concentrations. We know that the volumes of acid and base must sumto give 160 mL, or 0.160 L (use L because the unit M =mol/L). That is:     v(HA) = 0.160 - v(A-) Substituting this into the previous equation we have:   10-0.1 = (0.5*v(A-)) / (0.7*(0.160-v(A-)))             10-0.1 = (0.5*v(A-)) /(0.112-0.7*v(A-))         0.5*v(A-) =10-0.1(0.112-0.7*v(A-)) 0.5*v(A-) = 10-0.1*0.112 -10-0.1*0.7*v(A-))     10-0.1*0.112 = v(A-)(0.5 +10-0.1*0.7)               v(A-) = 10-0.1*0.112 / (0.5 + 10-0.1*0.7)                      = 0.0842 L And:      v(HA) = 0.160 - 0.0842                        = 0.0758 L Finally, we can check to see that these answers are correct:     n(A-) = 0.500 mol/L * 0.0842 L = 0.0421 mol    n(HA) = 0.700 mol/L * 0.0758 L = 0.05306 mol Henderson-Hasselbalch:     pH = 4.75 + log(0.0421 mol / 0.05306 mol)           = 4.65 =)

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