what is the concentration (molarity) of a solution that has 0.150 moles of sodium chloride dissolved in 0.900 L solution
1 answer:
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Required pH = 4.93
- OH⁻ from NaOH reacts with CH₃COOH giving CH₃COO⁻ and H₂O
- Let the volume of 3.5 M NaOH be x ml
Moles of NaOH = Moles of OH⁻ = Molarity * x ml = 3.5x mmol
- The reaction table for moles is as follows:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Initial 60 3.5x 40
Change -3.5x -3.5x +3.5x
Final (60-3.5x) 0 (40+3.5x)
- Substitute in Henderson equation and solve for x:
pH = pKa + log![\frac{[CH_{3}COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D%20)
4.93 = 4.76 + log
0.17 = log
x = 5.62 ml NaOH required
- OH⁻ from NaOH reacts with CH₃COOH giving CH₃COO⁻ and H₂O
- Let the volume of 3.5 M NaOH be x ml
Moles of NaOH = Moles of OH⁻ = Molarity * x ml = 3.5x mmol
- The reaction table for moles is as follows:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Initial 60 3.5x 40
Change -3.5x -3.5x +3.5x
Final (60-3.5x) 0 (40+3.5x)
- Substitute in Henderson equation and solve for x:
pH = pKa + log
4.93 = 4.76 + log
0.17 = log
x = 5.62 ml NaOH required
Answer:
The percentage is k %
Explanation:
From the question we are told that
The mass of the stibnite is
The volume of KBrO3(aq) is
The concentration of KBrO3(aq) is
Now the balanced ionic equation for this reaction is
The number of moles of is
substituting values
from the reaction we see that 1 mole of reacts with 3 moles of
so 0.003325 moles will react with x moles of
Therefore
Now the molar mass of is a constant with a values of
Generally the mass of is mathematically represented as
substituting values
The percentage of Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as
k
k %