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Calculate the sodium ion concentration when 70 mL of 3.0 M sodium carbonate is addred to 30.0 mL of 1.0 M sodium bicarbonate..

Im clueless

By Y17k (Y17k) on Tuesday, September 30, 2003 - 04:00 am: Edit

[Na+] in Sodium carbonate = 2 * 3.0 = 6.0 M

[Na+] in sodium bicarbonate = 1.0 M

[Na+] overall = 0.7 * 6.0 + 0.3 * 1.0

= 4.2 + 0.3

= 4.5 M

By Joel_Set (Joel_Set) on Tuesday, September 30, 2003 - 04:10 am: Edit

wow same here.. this question is the even question off of zhumdal.. is it 7.2? i tried and it doesn't really make sense.. anyone care to solve?

By Joel_Set (Joel_Set) on Tuesday, September 30, 2003 - 04:11 am: Edit

hey y17k isn't that the amount of moles.. not molarity?

By Joel_Set (Joel_Set) on Tuesday, September 30, 2003 - 04:14 am: Edit

and also what's the difference between molarity and concentration.. like i am so confused right now... i thought molarity and concentration were the same thing and

Molarity= moles/Volume

my teacher was talking some thing about like C= m x V.. i am really confused if amount of mole should be up there or is it the concentration.. i think that's where you got ur answer from.. plz respond

By Joel_Set (Joel_Set) on Tuesday, September 30, 2003 - 04:29 am: Edit

never mind.. is the answer really that simple? all u have to do is add the concentrations..?

By Y17k (Y17k) on Tuesday, September 30, 2003 - 05:17 am: Edit

Joel_set.

It works for molarity as well.

For example, 70 mls of 3.0 M of Na2CO3 would yield 0.21 MOLS, which would mean there are 0.42 MOLS of Na ions.

[Na+] = .42 * 1000/70 = 6.0 M

You COULD do this long process, but my way's easier ^^.

And your other question: it really is as simple as adding the concetration, as long as you follow the abundance weighted mean. I.E. you had 70 % of 6.0 M, and 30 % of 1.0 M, therefore the total molarity would be 0.7 * 6.0 + 0.3 * 1.0.

Hope thsu helps

By Joel_Set (Joel_Set) on Tuesday, September 30, 2003 - 05:48 am: Edit

k thx.. KP

By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:10 pm: Edit

OMG this question was on my test on monday. It was the last question and I got it wrong b/c I didn't multiply by 2. Darn! Too bad b/c I was rushing...

JOEL SET -- WHERE DID YOU GET THIS QUESTION?

"this question is the even question off of zhumdal" -- whats that?

By Special_Foreign (Special_Foreign) on Wednesday, October 01, 2003 - 06:46 pm: Edit

Y17k got the right answer,

3 M Na2CO3 = X/ .07 L
X =.21 mol Na2CO3 or .42 mol Na+ and .21 mol CO3-2

1 M NaHCO3 =X/.03 L
X= .03 mol NaHCO3 or .03 mol Na+ and .03 mol HCO3-

X=.45 mol Na+ ( this is from adding the .42 and .03 mol of Na+) / .1 L (this is from adding the volume of the two solutions)
X =4.5

4.5 M Na+

By Alimshk (Alimshk) on Wednesday, October 01, 2003 - 06:50 pm: Edit

I hate Chemistry . . . molarity/molality/concentration/stoichiometry etc. always confused me.