Question

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

Answer 1

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

Answer 2

Titration is a chemical process used to determine the concentration of an unknown sample from the known sample. At the equivalence point, the pH is 8.279.

What is the equivalence point?

In titration, the point at which the titrant neutralizes the unknown analyte solution is called the equivalence point.

The moles of equivalence point of sodium hydroxide and nitrous acid are:

$\begin{aligned}\rm n_{NaOH} &= \rm n _{HNO_{2}}\\\\&= 25.00 \times 0.200\\\\&= 0.005 \;\rm mol\end{aligned}$

The volume of the added base is 5.00 mL.

The moles of nitrogen dioxide are 0.005 moles and the volume at the equivalence point is 30 mL.

The molar concentration of nitrogen dioxide will be, 0.167 M.

From the ICE table, the base dissociation constant is calculated as:

$\begin{aligned}\rm K_{w} &= \rm K_{a} \times K_{b}\\\\\rm K_{b} &= \dfrac{1 \times 10^{-14}}{4.6 \times 10^{-14}}\\\\&= 2.174 \times 10^{-11}\end{aligned}$

The ionization constant of the base is given as,

$\begin{aligned}\rm K_{b} &= \rm \dfrac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]}\\\\2.174 \times 10^{-11} &= \rm \dfrac{x^{2}}{0.167- x}\\\\&= 1.905 \times 10^{-6} \end{aligned}$

Now, pOH is calculated as:

$\begin{aligned} \rm pOH &=\rm - log [OH^{-}]\\\\&= \rm -log (1.905 \times 10^{-6})\\\\&= 5.72\end{aligned}$

The pH of the titration is calculated as:

$\begin{aligned} \rm pH + pOH &= 14\\\\\rm pH &= 14.00 - 5.72\\\\&= 8.279\end{aligned}$

Therefore the pH is 8.27.

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