titration of a 12.0 mL solution of HCL requires 22.4 mL of 0.12 M NaOH. What is the molarity of the HCL solution?
2006-12-11 15:39:22 · 3 answers · asked by femylatina 1 in Science & Mathematics ➔ Chemistry
You used 0.0224 L of 0.12 mol/L solution, or 0.0224*0.12 = 0.002688 moles of NaOH. Each mole of NaOH neutralizes one mole of HCl, so there were 0.002688 moles of HCl in the solution. The solution had a volume of 0.012 L, so the concentration was 0.002688mol/0.012L, or 0.224 M.
2006-12-11 15:43:08 · answer #1 · answered by Amy F 5 · 0⤊ 0⤋
Molarity (M) is mols/Liters so if the balanced equation indicates one mole of NaOH is required for HCl than you simply:
1) Find the moles of NaOH: M = mol/L so mol-NaOH = M * L = .12 * 0.0224
2) Find the Molarity of the HCl solution: M = mol/L = Step1/0.012
2006-12-11 15:45:33 · answer #2 · answered by dgbaley27 3 · 0⤊ 0⤋
that's the position the M1xV1=M2xV2 equation kicks in (0.085M HNO3) x V1 = (0.12M Ba(OH)2) x 15mL Divide out the 0.0.5 Molar answer on the different area and also you need to get round 21.a million (remember significant figures!)
2016-11-25 22:06:22 · answer #3 · answered by ? 4 · 0⤊ 0⤋