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???
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
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.
To determine the volume of 0.101 M HNO3 required to neutralize 51.1 mL of 0.127 M NH3, you can use the concept of stoichiometry and the equation of the reaction between HNO3 and NH3.
The balanced equation for the reaction is:
HNO3 + NH3 -> NH4NO3
From the equation, we can see that 1 mole of HNO3 reacts with 1 mole of NH3 to form 1 mole of NH4NO3. Therefore, the stoichiometric ratio between HNO3 and NH3 is 1:1.
To find the volume of HNO3, we can set up a proportion using the molarity and volume relationship:
(Molarity of HNO3) x (Volume of HNO3) = (Molarity of NH3) x (Volume of NH3)
Let's plug in the given values:
(0.101 M) x (Volume of HNO3) = (0.127 M) x (51.1 mL)
Before solving, we must ensure that the units are consistent. Since the volumes are given in milliliters (mL), we need to convert them to liters (L) to match the molarity unit. There are 1000 mL in a liter, so:
(0.101 M) x (Volume of HNO3 in L) = (0.127 M) x (51.1 mL ÷ 1000 mL/L)
Now we can solve for the volume of HNO3:
Volume of HNO3 in L = (0.127 M) x (51.1 mL ÷ 1000 mL/L) ÷ (0.101 M)
Volume of HNO3 in L = (0.127 M) x (0.0511 L) ÷ (0.101 M)
The molarity of HNO3 cancels out, leaving us with:
Volume of HNO3 in L = 0.0511 L
Finally, we need to convert the volume back to milliliters:
Volume of HNO3 in mL = 0.0511 L x 1000 mL/L
Volume of HNO3 in mL = 51.1 mL
Therefore, the correct volume of 0.101 M HNO3 required to neutralize 51.1 mL of 0.127 M NH3 is 51.1 mL, not 193 mL.