mod/ibis/view.phpid 2608459 ity, San Bernardino . CHEM 345-Spring16 . NOBLET Act

Question

mod/ibis/view.phpid 2608459 ity, San Bernardino . CHEM 345-Spring16 . NOBLET Activities and Due Dates Gra 4/22/2016 12:00 PM78/100 Print Calalator End] Poriode Table Question 5 of 5 Incorrect Map 2.104 g of a solid mixture containing only potassium carbonate (FW 138.2058 g/mol) and potassium bicarbonate (FW-100 1154 grmol) is 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 is dissolved in distilled water. 31.44 mL of a 0.755 M HI standard K2CO KHCOa Number Number wt% wt % Tools Hint Previous Give up & View Souton 2, Check Answer Next Ext

Explanation / Answer

Let's write the reaction here:

HCO3- + H+ -------> H2O + CO2
CO32- + 2H+ ------> H2O + CO2

HCO3- + CO32- + 3H+ --------> 2H2CO3

This means that 1 mole of each component react with 3 moles of H+ so:
moles acid = 0.755 * 0.03144 = 0.0237 moles

moles HCO3- = 0.0237/3 = 0.0079 moles

I will call x to K2CO3 and y to KHCO3:

total mass = mx + my
2.104 = mx + my
2.104 = (nx * MWx) + (ny * MWy)
2.104 = (138.3058nx) + (100.1154ny) ---> dividing by 138.3058 we have:
2.104/138.3058 = nx + 100.1154ny/138.3058
0.0152 = nx + 0.7239ny
nx = 0.0152 - 0.7239ny
nx = 0.0152 - 0.7239(0.0079)
nx = 0.0095 moles

mx = 0.0095 * 138.3058 = 1.3139 g
my = 2.104 - 1.3139 = 0.7901 g

%K2CO3 = 1.3139 / 2.104) * 100 = 62.45%
%KHCO3 = 100 - 62.45 = 37.55%

Hope this helps

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