Solve and show work. Li2S + 2 HNO3 --> 2 LiNO3 + H2S (a) Calculate the mass of lithium sulfide that will react with 250 mL of
0.65 M nitric acid (b) How many mL of 9.6 M HNO3 are needed to make 250 mL of 0.65 M HNO3 solution? (c) How many grams of nitric acid are needed to make 750 mL of 9.6 M HNO3.
The question is incomplete. The complete question is :
In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.
Solution :
Consider the equation of the reaction between NaOH and
NaOH (aq) + HCl (aq) → NaCl(aq) +
The above equation tells us that of reacts with of .
So at the equivalence point, the moles of NaOH added = moles of present.
If the volume of the taken = mL and the conc. of = mole/L
The volume of NaOH added up to the color change = mole/L
Moles of taken = moles.
The color change will occur when the moles of NaOH added is equal to the moles of taken.
Thus when
or when
or mL of NaOH added, we observe the color change.
Where are the volume and molarity of the taken.
is the molarity of NaOH added.
When both the NaOH and are of the same concentrations, i.e. if , then
Or the 40 mL of will need 40 mL of NaOH for a color change and
30 mL of would need 30 mL of NaOH for the color change (provided the concentration )
1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is: