Question

1. a 30.00 ml sample of 0.100 m hocl ($k_a = 3.3\times10^{-8}$) is titrated with 0.100 m naoh.

a. write the balanced nie for the titration reaction.
b. calculation the expected ph at the equivalence point.
c. would thymol blue be a suitable indicator for this titration? explain your answer. (refer to your textbook and experiment 4.)

Explanation:

Step1: Write the balanced NIE

HOCl is a weak - acid and NaOH is a strong - base. The reaction is $HOCl(aq)+OH^-(aq)
ightarrow OCl^-(aq)+H_2O(l)$.

Step2: Calculate moles of HOCl

$n(HOCl)=M\times V = 0.100\ mol/L\times0.0300\ L = 0.00300\ mol$. At the equivalence point, the moles of NaOH added is equal to the moles of HOCl, and the volume of NaOH added $V(NaOH)=\frac{n(NaOH)}{M(NaOH)}=\frac{0.00300\ mol}{0.100\ mol/L}=0.0300\ L$. The total volume at the equivalence point $V_{total}=0.0300\ L + 0.0300\ L=0.0600\ L$. The concentration of $OCl^-$ formed $[OCl^-]=\frac{0.00300\ mol}{0.0600\ L}=0.0500\ mol/L$.

Step3: Set up Kb expression for $OCl^-$

$K_b=\frac{K_w}{K_a}=\frac{1.0\times10^{-14}}{3.3\times10^{-8}} = 3.03\times10^{-7}$. For the reaction $OCl^-(aq)+H_2O(l)
ightleftharpoons HOCl(aq)+OH^-(aq)$, let $x = [OH^-]=[HOCl]$, and $[OCl^-]=0.0500 - x$. Since $K_b$ is small, $0.0500 - x\approx0.0500$. Then $K_b=\frac{x\cdot x}{0.0500}$. So $x^2 = K_b\times0.0500=3.03\times10^{-7}\times0.0500 = 1.515\times10^{-8}$, and $x = [OH^-]=\sqrt{1.515\times10^{-8}}=1.23\times10^{-4}\ mol/L$.

Step4: Calculate pOH and pH

$pOH=-\log[OH^-]=-\log(1.23\times10^{-4}) = 3.91$. $pH = 14 - pOH=14 - 3.91 = 10.09$.

Step5: Evaluate the indicator

Thymol blue has a pH range of 8.0 - 9.6. Since the pH at the equivalence point is 10.09, which is outside the pH - range of thymol blue, thymol blue is not a suitable indicator for this titration.

Answer:

a. $HOCl(aq)+OH^-(aq)
ightarrow OCl^-(aq)+H_2O(l)$
b. 10.09
c. No, because the pH at the equivalence point (10.09) is outside the pH - range (8.0 - 9.6) of thymol blue.