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# What mass of potassium carbonate is needed to prepare 550. mL of an aqueous solution with a potassium concentration of 0.175M?

What mass of potassium carbonate is needed to prepare 550. mL of an aqueous solution with a potassium concentration of 0.175M?

Moles of K2CO3 needed = 0.175 x 0.550... View the full answer

• I got that answer too but my test review says the correct answer is 6.65g. Maybe because it's asking specifically for potassium?
• jazmin.garcia2000
• Nov 04, 2017 at 2:52am
• your question states potassium carbonate.. 1 mole of K2CO3 has 2 moles of potassium ions... hence if we have the moles of Potassium as 0.09625 moles, then the moles of the salt= 0.09625 /2=0.04813 moles... the mass= 0.04813 x 138.21= 6.651 g
• Astrobest2017
• Nov 04, 2017 at 3:04am
• hence 6.65 is a correct aanswer
• Astrobest2017
• Nov 04, 2017 at 3:04am
• since in the formula K2CO3 we have 2 moles of K per Mole of the salt, then we can just divide the mass 13.3 by 2 and we will arrive at the same answer of 6.65 g
• Astrobest2017
• Nov 04, 2017 at 3:11am

moles of potassium carbonate present in 550ml... View the full answer

• That's the answer I got as well but according to the test review, the correct answer is 6.65g. Any idea why that may be? P.S. Thank you for your response!
• jazmin.garcia2000
• Nov 04, 2017 at 2:51am
• considering the dissociation of aqueous potassium carbonate,it gives 1 mole of potassium ions and 1 mole carbonate ions;if we are to consider potassium ions only,then divide our original answer by 2; hence 13.2828/2=6.64125g as the correct answer
• ochiprof
• Nov 04, 2017 at 3:02am

we have volume of solution = 550 ml = 0.550 L and concentration of solution = 0.175 M... View the full answer

• concentration of potassium ions = 0.175 M
• william886088
• Nov 04, 2017 at 8:09am
• moles of potassium ions = molarity x volume = 0.175 M x 0.550 L = 0.09625 mole
• william886088
• Nov 04, 2017 at 8:10am
• but the dissociation of K2CO3 is given as : K2CO3 (aq) ---------------> 2 K+ (Aq) + CO3^2- (Aq)
• william886088
• Nov 04, 2017 at 8:10am
• Thus there will be , moles of K2CO3 = 0.09625 / 2 = 0.048125 mole
• william886088
• Nov 04, 2017 at 8:11am
• mass of potassium carbonate required = moles x molar mass = 0.048125 mole x 138.205 g/mol = 6.65 g
• william886088
• Nov 04, 2017 at 8:12am
• william886088
• Nov 04, 2017 at 8:12am

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