Question

1. To remove organic matter is one of the most critical tasks for water treatment. Low molecular-weight (LMW) organic acid is an important component of organic matter in the waste water. As a representative LMW organic acid, Citric acid, CoHs07 abbreviated as H3Cit, is a triprotic acid. With pkal-3.13, pKa2-4.72, pKa3-6.33. Consider a solution made by adding lemon juice to water until pH of the solution is 2.2. Assuming all the acidity is from dissociation of citric acid, find the total concentration of citrate species, TOTCit, and the concentration of HCit Citric acid 2. How much HOCI must be added to pure water to make a solution of pH 4.3? pH 6.5?

Answer 1

Chegg Policy: 1 questions per post or 4 sub parts in 1 question. According this I'm answering first question only

ANSWER:

System reactions:

$H_{3}Cit+H_{2}O\rightleftharpoons H_{2}Cit^{-}+H_{3}O^{+}\, \, \, \, \, pKa_{1}=3.13$

$H_{2}Cit^{-}+H_{2}O\rightleftharpoons HCit^{-2}+H_{3}O^{+}\, \, \, \, \, pKa_{2}=4.72$

$HCit^{-2}+H_{2}O\rightleftharpoons Cit^{-3}+H_{3}O^{+}\, \, \, \, \, pKa_{3}=6.33$

$2H_{2}O\rightleftharpoons H_{3}O^{+}+OH^{-}\, \, \, \, \, pKw=1x10^{-14}$

We need the values of Ka1, Ka2 and Ka3:

pka = -log ka Ka = 10-pka

$Ka_{1}=10^{-3.13}=7.41x10^{-4}$

Και = 10 72 1.91.r10)

$Ka_{3}=10^{-6.33}=4.68x10^{-7}$

The mass balance equation for citric acid solution is

$TOTCit=[H_{3}Cit]+[H_{2}Cit^{-}]+[HCit^{-2}]+[Cit^{-3}]$

and the balance charge equation is

$[H_{3}O^{+}]=[H_{2}Cit^{-}]+2[HCit^{-2}]+3[Cit^{-3}]+[OH^{-}]$

For a pH of 2.2, the [H3O+] is:

$pH=-log\left [ H_{3}O^{+}\right ]\, \, \, \, \, \Rightarrow \, \, \, \, \, [H_{3}O^{+}]=10^{-pH}$

Then, the [OH-] is

$Kw=[H_{3}O^{+}][OH^{-}]=1.00x10^{-14}\, \, \, \, \, \Rightarrow \, \, \, \, \,[OH^{-}]=\frac{Kw}{[H_{3}O^{+}]}$

$\mathbf{[OH^{-}]}=\frac{1.00x10^{-14}}{6.31x10^-3}=\mathbf{1.58x10^{-8}\, M}$

Using equilibrium constant expressions, we can express [H2Cit-], [HCit-2] and [Cit-3] as function of [H3Cit].

• [H2Cit-]

$Ka_{1}=\frac{[H_{2}Cit^{-}][H_{3}O^{+}]}{[H_{3}Cit]}=7.41x10^{-4}\, \, \, \, \, \Rightarrow \, \, \, \, \,[H_{2}Cit^{-}]=\frac{Ka_{1}[H_{3}Cit]}{[H_{3}O^{+}]}$

$\mathbf{[H_{2}Cit^{-}]}=\frac{7.41x10^{-4}[H_{3}Cit]}{6.31x10^-3}=\mathbf{0.117[H_{3}Cit]}$

• [HCit-2]

$Ka_{2}=\frac{[HCit^{-2}][H_{3}O^{+}]}{[H_{2}Cit^{-}]}=1.91x10^{-5}\, \, \, \, \, \Rightarrow \, \, \, \, \,[HCit^{-2}]=\frac{Ka_{2}[H_{2}Cit^{-}]}{[H_{3}O^{+}]}$

$\mathbf{[HCit^{-2}]}=\frac{1.91x10^{-5}\times 0.117[H_{3}Cit]}{6.31x10^-3}=\mathbf{3.54x10^{-4}[H_{3}Cit]}$

• [HCit-3]

$Ka_{3}=\frac{[Cit^{-3}][H_{3}O^{+}]}{[HCit^{-2}]}=4.68x10^{-7}\, \, \, \, \, \Rightarrow \, \, \, \, \,[Cit^{-3}]=\frac{Ka_{3}[HCit^{-2}]}{[H_{3}O^{+}]}$

$\mathbf{[Cit^{-3}]}=\frac{4.68x10^{-7}\times 3.54x10^{-4}[H_{3}Cit]}{6.31x10^-3}=\mathbf{2.63x10^{-8}[H_{3}Cit]}$

Ussing the balance charge equation, we can calculate the value of [H3Cit]:

$[H_{3}O^{+}]=[H_{2}Cit^{-}]+2[HCit^{-2}]+3[Cit^{-3}]+[OH^{-}]$

$6.31x10^{-3}\, M=0.117[H_{3}Cit]+2\times \left ( 3.54x10^{-4}[H_{3}Cit] \right )+3\times \left ( 2.63x10^{-8}[H_{3}Cit] \right )+1.58x10^{-8}\, M$

H3 Cit 0.0536 M

Now, ussing mass balance equation we can calculate the TOTCit:

$TOTCit=[H_{3}Cit]+[H_{2}Cit^{-}]+[HCit^{-2}]+[Cit^{-3}]$

$TOTCit=1.117[H_{3}Cit]=1.117\times 0.0536\, M$

$\mathbf{TOTCit=0.0599\, M}$

And the value of [HCit-2] is

$[HCit^{-2}]=3.54x10^{-4}[H_{3}Cit] = 3.54x10^{-4}\times 0.0536\, M$

$\mathbf{[HCit^{-2}]=1.90x10^{-5}\, M}$