In a particular titration experiment a 25.0 mL sample of an unknown diprotic acid required 30.0 mL of 200. M NaOH for the end point to be reached. What is the concentration of the acid?
0.120 M
0.167 M
0.240 M
0.333 M
Answer:
H2A + 2 NaOH >> Na2A + 2 H2O
Moles OH- = 30.0 x 0.200 /1000 = 0.006
The ratio between H2A and NaOH is 1 : 2
Moles H2A =0.006 / 2 = 0.003
Concentration acid = 0.003 moles / 0.025 L = 0.120 M
for example:
H2SO4 + 2NaOH --->Na2SO4 + 2H2O
n(a) = n(b)/2
c(a)V(a) = c(b)V(b) / 2
c(a) = c(b)V(b) / 2V(a)
c(a) = 0,2 M × 30ml / (2×25ml)
c(a) = 0,12 M
c(a) concentration of the acid
c(b) concentration of the base (NaOH)
diprotic acids have 2H+ ions. The general form is H2X, where X can be, for instance, SO4, as it is used above.
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