That is the symbol that depicts an oxidizing reagent
Answer:
28.58 g of NaOH
Explanation:
The question is incomplete. The missing part is:
<em>"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"</em>
To do this, we need to know how much of the base we have to weight to prepare this solution.
First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:
NaOH -------> Na⁺ + OH⁻
As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.
This can be done with the following expression:
14 = pH + pOH
and pOH = -log[OH⁻]
So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:
pOH = 14 - 13.9 = 0.10
[OH⁻] = 10⁽⁻⁰°¹⁰⁾
[OH⁻] = 0.794 M
Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:
n = M * V
n = 0.794 * 0.9
n = 0.7146 moles
Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:
m = 39.997 * 0.7146
<h2>
m = 28.58 g</h2>
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Answer:
pH of soltion will be 5.69
Explanation:
The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.
The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:
![pH=pKa+log(\frac{[salt]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%20%29)
The moles of acid taken :
The moles of base taken:
The moles of acid left after reaction :
The moles of salt formed = 4.5mmol
Putting values in equation
![pH=pKa+log(\frac{[salt]}{[acid]} )=4.74+log(\frac{4.5}{0.5})=5.69](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%20%29%3D4.74%2Blog%28%5Cfrac%7B4.5%7D%7B0.5%7D%29%3D5.69)
