moles H2X present = 0.1873/85.00 = ??
moles NaOH = 2 x that number.
MNaOH = moles NaOH/L NaOH
Solve for L NaOH and convert to mL. 41.89 mL is the correct answer.
A 0.1873 g sample of a pure, solid acid, H2X was dissolved in water and titrated with 0.1052 M NaOH solution.
The balanced equation for the neutralization reaction occurring is
H2X(aq) + 2NaOH(aq) �¨ Na2X(aq) + 2H2O(l)
If the molar mass of H2X is 85.00 g/mol, calculate the volume of NaOH solution needed in the titration. (41.89 mL)
3 answers
How do I solved for liters if nothing is given in the question?
What do you mean by "nothing" is given in the question. There is plenty given in the question. The problem gives grams and molar mass and from that you calculate moles. You look at the balanced equation, multiply by 2 and that gives you moles NaOH. Then you solve M = moles/L (rearranged to L = moles/M) and solve for L, then convert to mL. I solved the problem for you; the only thing I didn't do is to work out the math.