a neutral solution?
2007-05-16 15:10:43 · 3 answers · asked by yowutsup911 1 in Science & Mathematics ➔ Chemistry
mL HCl = (MmL KOH* stoichiometric ratio) / M HCl
=(1.00M * 25.0mL * 1 HCl/1KOH) / 0.45M
=2.2
2007-05-17 14:23:50 · answer #1 · answered by zanekevin13 4 · 0⤊ 2⤋
Since it's a strong acid and a strong base, you need equal amounts of OH- and H+ ions. Both will dissociate completely. So it becomes a pretty simple problem, you just have to calculate how many mmols or mols of each you have. Since you have 25 ml of 1 M KOH, you have .025 mol KOH. Your HCl is .45 M. How many mL HCl would equal .025 mol HCl? It's about half the strength, so you would expect to need roughly twice as much. Do the math and you'll have your answer.
2007-05-16 15:22:42 · answer #2 · answered by some girl 3 · 0⤊ 0⤋
.
HCL (aq) + KOH (aq) → H2O (l) + KCl (aq)
25.0 mL KOH = .0250 L KOH x (1.00 mol) / 1L = 0.0250 mol KOH
0.025 mol KOH x [(1 mol HCl)/(1 mol KOH)] x [(1 L HCl)/(0.45 mol HCl)] = .0556 L HCl
2016-05-24 16:15:39 · answer #3 · answered by Dennis 1 · 1⤊ 0⤋