Answer to Question #155460 in Inorganic Chemistry for Salma

Question

 

A sample, KOH, K2CO3, dissolved in water (35cm3). Its solution was titrated with 1.058moldm-3 HCl requiring 17.7cm3 to make it colourless due to phenolphthalein solution. After adding methyl orange, the solution consumed further 6cm3 of the acid to complete the neutralisation. What is the concentration of the sample?

Answer 1

$2HCl + K_2CO_3 \to 2KCl + CO_2 + H_2O\\ HCl + KOH \to KCl + H_2O$

let;

a represent hcl

b and c represent k2co3 and koh respectively

c_a = 1.058 mol/dm³

v_a for k2co3 = 17.7cm³

v_a for koh = 6cm³

v_b =35cm³

v_c = 35cm³

Using cava/cbvb = na/nb

1.058×17.7/cb×35 = 2/1

cb = 0.268 mol/dm³

Using cava/ccvc = na/nc

1.058×6/cc×35 = 1/1

c_c = 0.181 mol/dm³