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?
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"
Comments
Leave a comment