Answer to Question #155459 in Inorganic Chemistry for Salma

Question

An alkaline sample of sodium compounds (NaOH, Na2CO3) was dissolved in water 25cm3, was titrated with 1.058moldm-3 H2SO4 at 6.8cm3 to make it the solution colourless after phenolphthalein was added. After adding methyl orange a further 16.5cm3 more of the acid were needed to complete the analysis. What is the concentration of both?

Answer 1

Since there are two sodium compounds mixed in one sample, we have to define their concentrations step by step depending on their nature. The sodium hydroxide ($NaOH$) is a strong base and will react with $H_2SO_4$ in the following way:

$2NaOH + H_2SO_4 = Na_2SO_4+2H_2O$

During the reaction, the base is neutralizing by the acid and if we have phenolphthalein as an indicator, its colour will disappear after the neutralization.

Thus, the concentration of $NaOH$ can be obtained in a way:

$n(H_2SO_4) = n(NaOH)$

$c(H_2SO_4)\times V(H_2SO_4) = c(NaOH) \times V(NaOH)$

$c(NaOH) = \frac{c(H_2SO4)\times V(H_2SO_4)}{V(NaOH)}=\frac{1.058 mol/dm^3\times6.8 cm^3}{25.0 cm^3}=0.288 mol/dm^3$

Since the sodium bicarbonate ($Na_2CO_3$) is present in the sample, during the titration by $H_2SO_4$ methyl orange will change the colour to red showing the neutralization is complete:

$Na_2CO3 + H_2SO_4 = Na_2SO_4 +CO_2+H_2O$

The concentration of $Na_2CO_3$ can be computed by the similar scheme:

$n(H_2SO_4) = n(Na_2CO_3)$

$c(H_2SO_4)\times V(H_2SO_4) = c(Na_2CO_3) \times V(Na_2CO_3)$

$c(Na_2CO_3) = \frac{c(H_2SO4)\times V(H_2SO_4)}{V(Na_2CO_3)}=\frac{1.058 mol/dm^3\times 16.5 cm^3}{25.0 cm^3}= 0.698 mol/dm^3$