Answer:
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Explanation:
Explanation:
Reaction equation is as follows.

Here, 1 mole of
produces 2 moles of cations.
![[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58](https://tex.z-dn.net/?f=%5BNa%5E%7B%2B%7D%5D%20%3D%202%5BNa_%7B2%7DSO_%7B3%7D%5D%20%3D%202%20%5Ctimes%200.58)
= 1.16 M
= 0.58 M
The sulphite anion will act as a base and react with
to form
and
.
As, 
= 
=
According to the ICE table for the given reaction,

Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
![K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHSO%5E%7B-%7D_%7B3%7D%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BSO%5E%7B2-%7D_%7B3%7D%5D%7D)


x = 0.0003 M
So, x =
= 0.0003 M
= 0.58 - 0.0003
= 0.579 M
Now, we will use
= 0.0003 M
The reaction will be as follows.

Initial: 0.0003
Equilibrium: 0.0003 - x x x


= 
= 
Therefore, 
As, x <<<< 0.0003. So, we can neglect x.
Therefore, 
= 
x = 
x =
= 
![[H^{+}] = \frac{10^{-14}}{[OH^{-}]}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B%5BOH%5E%7B-%7D%5D%7D)
= 
=
M
Thus, we can conclude that the concentration of spectator ion is
M.
Answer:
Explanation:
The result will be affected.
The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).
Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.
During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.
Thus, NaOH will appear at a higher concentration than it actually is.