a) If 2.40 g NaHCO3 is added to 150 mL of 0.200 M HCl, calculate the molarity of H+ after the reaction above goes to completion.
b) What volume, in mL, of 0.900 M KOH would be required to neautralize the remaining HCl in the problem above?
This is a limiting reagent problem and the problems suggests that the NaHCO3 is the LR.
mols NaHCO3 = grams/molar mass = ?
mols HCl = M x L = ?
NaHCO3 + HCl ==> NaCl + CO2 + H2O
mols HCl remaining = initial mols HCl - mols NaHCO3 = ?
Then M HCl remaing - mols HCl/L solution. This assume that all of the CO2 gas has left the solution and that none of the gas is dissolved.
This is what I've done so far:
2.40 g NaHCO3 / 84.01 g = 0.0285 mol NaHCO3
0.200 M * 0.150 L = 0.0300 mol HCl
0.0300 mol HCl - 0.0285 mol NaHCO3 = 0.0015 mol HCl remaining
Now what? Which values do incorporate in the M of HCl?
I don't get the "then M HCl remaining - mols HCl/L soln" which numbers do I use here?:
You've done the subtraction step and the and 0.0015 mols HCl is what is remaining. All you need to do now is to convert that to M = mols/L which is 0.0015/total volume in L = ?
what is the total volume that i should use. which value given?
nevermind, i got it. now the second part, how do i start?
There is only one volume given and that is 150 mL or 0.150 L.
For the other one,
KOH + HCl ==> KCl + H2O
How many mols HCl remains? That is 0.00150 mols.
The equation tells you that 1 mol KOH is needed for 1 mol HCl; therefore, you must use 0.0015 mols KOH. Molarity = mols/L. You have mols KOH and you have M KOH in the problem. Solve for volume in L and convert to mL.
A note: I sense that you are overwhelmed with chemistry. I have worked several problems of the same type for you, yet little information appears to be transferred from one problem to the next. You fail to pick up volumes and molarities that are in the problem. You get the answer but don't recognize that's the answer. I suggest you get a tutor to help you one on one. I am happy to continue helping but my efforts don't seem to be enough; I think you need more one to one help in a face to face manner. You need to remember that reactions are between mols. If you have a solid then mols = grams/molar mass. If you have a liquid mols = M x L. The third important ingredient is you need to be able to convert mols of one material in the equation to any other material in the reaction by using the corresponding coefficients in the balanced equation. Good luck. By the way, your post is now far down the list and I suggest you repost at the top if you have further comments about this problem.