2.180 g of a solid mixture containing only potassium carbonate (FW = 138.2058 g/

Question

2.180 g of a solid mixture containing only potassium carbonate (FW = 138.2058 g/mol) and potassium bicarbonate (FW = 100.1154 g/mol) is dissolved in distilled water. 35.00 mL of a 0.785 M HCl standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture.

Mapddo sapling learning 2.180 g of a solid mixture containing only potassium carbonate (FW = 138.2058 g/mol) and potassium bicarbonate (FW = 100.1 154 g/mol) is dissolved in distilled water. 35.00 mL of a 0.785 M HCl standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture K2CO3 KHCO3 Number Number click to edit wt% wt %

Explanation / Answer

1 mol of K2CO3 will require 2 mol of HCl and 1 mol of KHCO3 will require 1 mol of HCl

Let mass of K2CO3 be x gm then mass of KHCO3 = (2.18-x ) g

Molar mass of K2CO3= 138.2 g/mol
number of moles of K2CO3 = mass/molar mass = x/138.2 mol

Molar mass of KHCO3= 100.1 g/mol
number of moles of KHCO3 = mass/molar mass = (2.18-x)/100.1 mol

so, number of moles of HCL required = 2*x/138.2 +(2.18-x)/100.1
                                                                              = 14.47*10^-3 x + 0.022 -9.99*10^-3x
                                                                              = 4.482*10^-3*x + 0.022 mol

Actually moles of HCl added =M*V = 0.785 M*0.035 L = 0.0275
SO,
4.482*10^-3* x + 0.022 = 0.0275
x = 1.23
mass % K2CO3= X*100/2.18 = 1.23*100/2.18 = 56.3 %
Mass % of KHCO3= 100 - 56.3 = 43.7 %

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.