Asked by lyne

What volume of .101 M HNO3 is required to neutralize each of the following solutions?

there are four and i got them all right but this one...

51.1 mL of .127 M NH3

i got 193 mL

can someone please help???
Answered by DrBob222
Surely, but why don't you show your work and we can help find the error.
Answered by lyne
.0511 L NH3 x (.127 mol NH3/L NH3) x (3mol HNO3/1 mol KOH) x (L KOH/.101 ml HNO3) = .193L = 193 mL

i don't know where i messed up
Answered by DrBob222
The problem you posted doesn't say anything about KOH.
The problem you posted says 51.1 mL of 0.127 M NH3 is neutralized with 0.101 M HNO3. So you omit the 3 mol HNO3/1 mol KOH) and 1 L KOH/0.101 mL and replace them with 1 L HNO3/0.101 mols) = ??
Check my work.
I usually work these by
mL x M = mL x M.
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