Neutralization equation:
Moles acid= 2*moles Base (balance the equation to see the 2)
.2/150=2*.15*Vb(in liters)
Vb=.2/(150*2*.15) =.004 liters check that
2. Suppose you measured out 0.20 g of tartaric acid, a diprotic acid, H2C4H4O6 , determine the volume of 0.15 M sodium hydroxide required to neutralize this acid.
2 answers
This is a diprotic weak acid with a pKa₁ of 2.89 => Ka₁ = 1.3 x 10⁻³ and a pKa₂ = 4.40 => Ka₂ = 4.0 x 10⁻⁵. With such a low Ka₂, it is assumed (generally) that all of the Hydronium ions come from the 1st Ionization Step.
For 0.20-g H₂Tartarate (=0.2/150 mole) => 0.0013-mole Tartaric Acid being added into solution.
Assuming a 1-Liter solution => 0.0013M Tartartic Acid (Ka₁ = 1.3 x 10⁻³) delivers ~ 0.0013M H₃O⁺ in solution.
Assuming all Hydronium ions are from the 1st ionization step & using the ICE Table analysis => [H₃O⁺] = SqrRt(Ka∙[Tartaric Acid]) = SqrRt((1.3 x 10⁻³)²) = 0.0013M H₃O⁺ => 0.0013-mole/Liter, needs neutralizing.
Molarity x Volume = moles => Volume(L) = moles/Molarity = 0.0013-mole/0.15-M = 0.0087-L = 8.7-ml 0.15M NaOH(aq).
For 0.20-g H₂Tartarate (=0.2/150 mole) => 0.0013-mole Tartaric Acid being added into solution.
Assuming a 1-Liter solution => 0.0013M Tartartic Acid (Ka₁ = 1.3 x 10⁻³) delivers ~ 0.0013M H₃O⁺ in solution.
Assuming all Hydronium ions are from the 1st ionization step & using the ICE Table analysis => [H₃O⁺] = SqrRt(Ka∙[Tartaric Acid]) = SqrRt((1.3 x 10⁻³)²) = 0.0013M H₃O⁺ => 0.0013-mole/Liter, needs neutralizing.
Molarity x Volume = moles => Volume(L) = moles/Molarity = 0.0013-mole/0.15-M = 0.0087-L = 8.7-ml 0.15M NaOH(aq).
3 answers