What volume, in liters, of 1.00 M KOH solution should be added to a 0.118 L solution containing 10.32 g of glutamic acid hydrochloride (Glu, FW = 183.59 g/mol; pKa1 = 2.23, pKa2 = 4.42, pKa3 = 9.95) to get to pH 10.31?
Moles of glutamic acid HCl present = mass/MW = 10.32/183.59 = 0.0562 moles
Since glutamic acid HCl is a weak acid, only its first dissociation is significant. Thus we will consider only the pKa1 value, because the other two dissociations will be insignificant.
Assume the volume of KOH taken as 'x' Liters.
Thus, moles of KOH taken = Volume*Molarity = x*1 = x moles
When 'x' moles KOH are added, 'x' moles of glutamic acid HCl react with them to form 'x' moles of salt.
Thus, moles of salt formed = x
Moles of acid left = 0.0562 - x
According to Henderson-Hasselbach equation,
pH = pKa + log(moles of salt/moles of acid)
10.31 = 2.23 + log(x/0.0562-x)
Solving, we get : x = 0.0562 L approximately
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