# Chemistry Damon

1g of mixture of na2c03 and k2c03 was made upto 250ml is aqueous solution . 25ml of this solution was neutralized 20ml of hcl of unknow concentration. The neutralized solution required 16.24ml of 0.1N AGN03 for precipitation. Calculate a) the k2c03 is mixture b) conc of hcl in g/L C) molarity of hcl

Damon wants you step by step for me.

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1. Na2CO3 =2(23)+12+3(16) = 106 g/mol
x moles of Na2 CO3 mass = 106 x grams

K2CO3 = 2(39)+12+3(16) = 138 g/mol
y moles of K2CO3 mass = 138 y grams

106 x + 138 y = 1 gram total

in the titration I used 25 mL of my 250 ml so I used 1/10 gram solids of my original carbonates mixture

The reactions are
Na2CO3 + 2HCl --> 2NaCl + H2O + CO2
K2CO3 + 2HCl --> 2 KCl + H2O + CO2

Not being a chemist I will assume that I get the same number of moles of NaCl plus KCl as I had moles of AgNO3
How many is that?
moles AgNO3 = .1moles/liter * .01624 L = .001624 moles of NO3-
so I had .001624 moles of NaCl plus KCl
which means I had .000812 moles of
Na2CO3 plus K2CO3
x/10 moles Na2CO3 + y/10moles K2NO3 =.000812 moles
or
x + y = .00812
and from way long ago
106 x + 138 y = 1 gram total
---------------------
106 (.00812-y) + 138 y = 1

.861 -106 y + 138 y = 1
32 y = .139
y = .00435
fraction of y which is K2CO3 = .00435/.00812 = .536 which is 53.6 %

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posted by Damon
2. Wait a minute, x and y are moles and you want percentage of grams perhaps.
x = .00377
108 x = .40716
138 y = .6003

so yes, 60% K2CO3

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posted by Damon

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