Question

Bromocresol purple is an indicator that changes color over a range from ph=5.2 to ph=6.8. what is ka of bromocresol purple?

Answer 1

The reaction during the transition point is:

HInd ⇄ H⁺ + Ind⁻

Such that

Ka = [H⁺][Ind⁻]/[HInd]

At transition point, HInd = Ind⁻, so what's left is H⁺. We determine H⁺ from the midpoint of the pH range given. The solution is as follows:

Midpoint = (5.2+6.8)/2 = 6
pH = -logH⁺
6 = -logH⁺
H⁺ = 1×10⁻⁶ = Ka

Answer 2

The  ${{\text{K}}_{\text{a}}}$ of bromocresol purple is $\boxed{{\text{0}}{\text{.000001}}}$.

Further explanation:

Indicator:

An indicator is any substance that indicates the presence or absence of a chemical species such as an acid or an alkali in the solution, usually by a color change. Methyl orange, methyl yellow, thymol blue, bromocresol purple are some examples of indicators.

Bromocresol purple (BCP) is an example of pH indicator and is a dye of the triphenylmethane family. It is also known as 5’, 5”-dibromo-o-cresolsulfophthalein. It is yellow below pH 5.2 and its color becomes violet above pH 6.8.

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

${\text{pH}}=-{\text{log}}\left[{{{\text{H}}^ + }}\right]$                                                              …… (1)

Here,

$\left[{{{\text{H}}^ + }}\right]$ is hydrogen ion concentration.

The general dissociation reaction of BCP is as follows:

 ${\text{HBCP}}\rightleftharpoons{{\text{H}}^ + }+{\text{BC}}{{\text{P}}^ - }$

Here,

HBCP is the indicator in undissociated form.

${{\text{H}}^ + }$ is hydrogen ion.

${\text{BC}}{{\text{P}}^ - }$ is the indicator in ionized form.

The formula to calculate the equilibrium constant of HBCP is as follows:

${{\text{K}}_{\text{a}}}=\dfrac{{\left[ {{{\text{H}}^ + }}\right]\left[ {{\text{BC}}{{\text{P}}^ - }} \right]}}{{\left[{{\text{HBCP}}}\right]}}$      ...... (2)

Here,

 ${{\text{K}}_{\text{a}}}$is the dissociation constant of HBCP.

 [HBCP] is the concentration of undissociated bromocresol purple.

 $\left[ {{{\text{H}}^ + }}\right]$ is hydrogen ion concentration.

 $\left[{{\text{BC}}{{\text{P}}^ - }}\right]$ is the concentration of dissociated bromocresol purple.

Rearrange equation (1) to calculate $\left[ {{{\text{H}}^ + }}\right]$ .

$\left[ {{{\text{H}}^ + }}\right]={10^{ - {\text{pH}}}}$       ...... (3)                                                                     …… (3)

Bromocresol purple is an indicator whose pH ranges from 5.2 to 6.8. So its mid pH can be calculated as follows:

$\begin{aligned}{\text{pH}}&=\dfrac{{5.2 + 6.8}}{2}\\&=\frac{{12}}{2}\\&=6\\\end{aligned}$

So the pH of BCP is 6.

Substitute 6 for pH in equation (3).

$\begin{aligned}\left[ {{{\text{H}}^ + }}\right]&={10^{ - 6}}\\&=0.000001\;{\text{M}}\\\end{aligned}$

 

At the endpoint,  [HBCP] becomes equal to $\left[{{\text{BC}}{{\text{P}}^ - }}\right]$ . So equation (2) becomes,

${{\text{K}}_{\text{a}}}=\left[ {{{\text{H}}^ + }}\right]$                                      …… (4)

Substitute 0.000001 M for $\left[{{{\text{H}}^ + }}\right]$ in equation (4).

 ${{\text{K}}_{\text{a}}}=0.000001$

So the value of ${{\mathbf{K}}_{\mathbf{a}}}$  for bromocresol purple is 0.000001.

Learn more:

1. Which indicator would be the best for the given titration? brainly.com/question/9236274

2. Example of physical change: brainly.com/question/1119909

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Mixture

Keywords: pH, ka, indicator, BCP, bromocresol purple, H+, 0.000001, 0.000001 M, 6, 5.2, 6.8, dissociation constant, undissociated, dissociated.