Answer : The
ion concentration is,
and the pH of a buffer is, 2.95
Explanation : Given,

Concentration of
(weak acid)= 0.26 M
Concentration of
(conjugate base or salt)= 0.89 M
First we have to calculate the value of
.
The expression used for the calculation of
is,

Now put the value of
in this expression, we get:



Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKNO_2%5D%7D%7B%5BHNO_2%5D%7D)
Now put all the given values in this expression, we get:


The pH of a buffer is, 2.95
Now we have to calculate the
ion concentration.
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![2.95=-\log [H_3O^+]](https://tex.z-dn.net/?f=2.95%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1.12\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.12%5Ctimes%2010%5E%7B-3%7DM)
The
ion concentration is, 
<span>Group 1 can be characterized as atoms that have 1 electron in their valence shell. This is valuable when dealing with these questions, because the loss or gain of valence electrons is what defines ionic relationships. When group 1 elements form ionic bonds with other atoms, they are extremely likely to lose their valence electron, since the nucleus has a weaker pull on it than, say, a chlorine atom has on its 7 valence electrons. The weaker pull between the nucleus and the valence electron of group 1 elements means that the radius is high, since the electron is more free to move with less pull on it. This also means that the first ionization energy is low, since it takes relatively little energy for that electron to be pulled away to another atom.</span>
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.
Balanced Equation: CuCl + NaNO₃ → NaCl + CuNO₃
Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol
the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,
then moles of NaCl = 0.31 mol
Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g
⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.
Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %
NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.
The correct answer is c, it’s evaporating there for it’s a chemical change