Can you help me do this one? Thanks:
4. A 2.34 mol/L sodium hydroxide solution is reacted with 11.6 g of lead (II) sulphate powder. What volume of sodium hydroxide would react completely?
And one more question, would you multiply the answer by 4 since it is 4OH- for every PbSO4? Thanks.
PbSO4 + 2NaOH ==> Pb(OH)2 + Na2SO4 or
PbSO4 + 4NaOH ==> Na2Pb(OH)4 + Na2SO4
Equation 1 if the reaction stops at Pb(OH)2 or equation 2 if the Pb(OH)4^2- complex ion is formed.
11.6 g PbSO4/molar mass PbSO4 = ?? mols PbSO4.
Mols NaOH required = mols PbSO4 x 2 (or 4 depending upon which reaction you are working with) = mols NaOH.
L NaOH = mols NaOH/molarity NaOH.
I hope this helps.
Ok, so this is what I did:
Step #1:
11.6 g PbSO4/ 303.28 molar mass PbSO4 = 0.038 mols PbSO4
Step #2:
0.038 mols PbSO4 x 4 = 0.153
Step #3:
0.153 x (2/4)= 0.0765
ratio
Step #4:
0.0765 x (1/2.34)=0.033L
Would this be right?
Ok, so this is what I did:
Step #1:
11.6 g PbSO4/ 303.28 molar mass PbSO4 = 0.038 mols PbSO4 There are three significant figures in 11.6 g PbSO4; therefore, the answer should be carried out to at least one more place. Your next answer checks so I think you kept that number in your calculator but just didn't type it into the answer.
Step #2:
0.038 mols PbSO4 x 4 = 0.153 This answer is ok BUT it is mislabeled. The 0.03825 mol is mol PbSO4. That times 4, then, is mol NaOH. That is
0.03825 mol PbSO4 x (4 mol NaOH/1 mol PbSO4) [from the equation]= 0.153 mol NaOH. Note how the units of mol PbSO4 cancel to leave the unit of mol NaOH (and that is the unit you want to proceed to the next step of liters NaOH).
Step #3:
0.153 x (2/4)= 0.0765
ratio I am confused as to why you want the ratio of the two equations unless the question is somehow related to the ratio. However, as originally stated, the ratio is not involved. You pick whichever reaction is involved and go with it. Step two changed from mols PbSO4 to mols NaOH and we need not make another change here.
Step #4:
0.0765 x (1/2.34)=0.033L
So this step is 0.153 mols NaOH x (1/2.34) = ?? liters NaOH.
Let me know if this isn't clear but if you repost be sure to explain exactly what you have trouble understanding.
Your calculation in Step #4 is correct, assuming that the molarity of the sodium hydroxide solution is 2.34 mol/L. So, based on your calculations, the volume of sodium hydroxide solution required to react completely with 11.6 g of lead (II) sulphate is approximately 0.033 L or 33 mL.
Regarding the second question about multiplying the answer by 4, it seems like you are referring to the stoichiometric ratio in the balanced chemical equation. If the reaction goes to completion according to Equation 2, then 4 moles of sodium hydroxide are required for each mole of lead(II) sulphate. However, in the calculation, you have already accounted for this by multiplying the molar quantity of lead(II) sulphate by 4 in Step #2. So, there's no need to multiply the final answer by 4.
Just follow the given calculation steps, and you should arrive at the correct answer. If you have any further questions or need clarification, feel free to ask.