Question

25mL of 0.107M ${H_3}P{O_4}$ was titrated with 0.115M solution of NaOH to the endpoint identified by indicator bromocresol green. This required 23.1mL of the 0.115M NaOH. The titration was repeated using phenolphthalein as an indicator. This time 25mL of 0.107M ${H_3}P{O_4}$required 46.2mL of the 0.115M NaOH. What is the coefficient of n in this equation for each reaction?
${H_3}P{O_4} + nO{H^ - } \to n{H_2}O + {\left[ {{H_{3 - n}}P{O_4}} \right]^{n - }}$

Answer 1

Hint: At first, we’ll calculate the number of milliequivalents of${H_3}P{O_4}$ and then number of milliequivalents of $O{H^ - }$ ions used in both the titrations. Then, we’ll equate milliequivalents of ${H_3}P{O_4}$ and $O{H^ - }$ions in the first titration and then second titration. The number of milliequivalents is equal to the product of volume, concentration and acidity.

Complete step by step answer:
 In a titration, 25mL of 0.107M ${H_3}P{O_4}$ (Phosphoric acid) was mixed with 0.115M solution of NaOH (sodium hydroxide) which is identified by an indicator bromocresol green and 23.1mL of NaOH is required in this. So, now we’ll calculate the number of milliequivalents (mEq) of $O{H^ - }$ions. A milliequivalent is equal to the number of univalent ions $\left( {{H^ + }\,or\, O{H^ - }} \right)$ which will be needed to react with one molecule of the substance. Now, we’ll calculate the number of milliequivalents of ${H_3}P{O_4}$ and let it be represented by m. Thus, m = volume × concentration × acidity(n)
$m = 25 \times 0.107 \times n$
$m = 2.675n$
Now, in the first titration, ${m_1} = 23.1 \times 0.115 \times 1$ [acidity of NaOH is 1 and number of milliequivalents of $O{H^ - }$ions used in this titration be ${m_1}$ ]
${m_1} = 2.656$
Hence, $2.675n = 2.656$
$n = \dfrac{{2.656}}{{2.675}}$ i.e., 1 (rounded off)
Now, this titration was done using phenolphthalein as an indicator and let the number of milliequivalents of $O{H^ - }$ ions used be ${m_2}$ .
${m_2} = 46.2 \times 0.115 \times 1$
${m_2} = 5.313$
So,$2.675n = 5.313$
$n = \dfrac{{5.313}}{{2.675}}$ i.e., 2 (rounded off)

Note: Remember the formula of number of milliequivalents i.e., the product of volume, concentration and acidity and acidity of $O{H^ - }$ions used is 1. Also, the number of milliequivalents doesn’t have any unit as the unit of volume is millilitre, concentration is moles per litre. So, the units get cancelled.