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orchidee7
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Please post homework and homework like problems in the homework forum
I'm stuck on how to find how many ml of NaOH were used in a back titration.
A 0.6000 g sample of K2CO3 (138.2055 g/mol) is dissolved in enough water to make 200.0 mL of solution A. A 20.00 mL aliquot of solution A is taken and put into an Erlenmeyer flask. To the flask is added 20.00 mL of 0.1700 M HCl:
K2CO3(aq) + 2HCl(aq)2KCl(aq) + H2O(l) + CO2(g)
The resulting solution is then titrated with 0.1048 M NaOH.
NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)
How many mL of NaOH are used? (24.16)
I have:
0.6000gK2CO3 x (1/138.2055gK2CO3) x (1/.200ml) = 0.021707 M K2CO3.
0.170M HCl x (1/0.04ml) = 0.0034 mol HCl and 0.0034 mol NaOH (1:1 ratio)
From here I don't know what to do to find the ml of NaOH required to solve the equation. Could anybody show me how or give me some hints? Thank you.
A 0.6000 g sample of K2CO3 (138.2055 g/mol) is dissolved in enough water to make 200.0 mL of solution A. A 20.00 mL aliquot of solution A is taken and put into an Erlenmeyer flask. To the flask is added 20.00 mL of 0.1700 M HCl:
K2CO3(aq) + 2HCl(aq)2KCl(aq) + H2O(l) + CO2(g)
The resulting solution is then titrated with 0.1048 M NaOH.
NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)
How many mL of NaOH are used? (24.16)
I have:
0.6000gK2CO3 x (1/138.2055gK2CO3) x (1/.200ml) = 0.021707 M K2CO3.
0.170M HCl x (1/0.04ml) = 0.0034 mol HCl and 0.0034 mol NaOH (1:1 ratio)
From here I don't know what to do to find the ml of NaOH required to solve the equation. Could anybody show me how or give me some hints? Thank you.