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???
3 answers
Surely, but why don't you show your work and we can help find the error.
.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
i don't know where i messed up
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.
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.