Answer:
32 mL
Explanation:
<em>A chemist must prepare 500.0mL of hydrobromic acid solution with a pH of 0.50 at 25°C. He will do this in three steps: Fill a 500.0mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (5.0M) stock hydrobromic acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrobromic acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
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Step 1: Calculate [H⁺] of the dilute solution
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -0.50 = 0.32 M
Step 2: Calculate [HBr] of the dilute solution
HBr is a strong acid that dissociates according to the following equation.
HBr ⇒ H⁺ + Br⁻
The molar ratio of HBr to H⁺ is 1:1. The concentration of HBr is 1/1 × 0.32 M = 0.32 M.
Step 3: Calculate the volume of the concentrated HBr solution
We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.32 M × 500.0 mL / 5.0 M
V₁ = 32 mL
Answer:
Cd(HCO₃)₂ (s) ⇒ CdCO₃ (s) + H₂O (l) + CO₂ (g)
Explanation:
Cadmium hydrogen carbonate = Cd(HCO₃)₂ (solid)
Cadmium carbonate = CdCO₃ (solid)
Water = H₂O (liquid)
Carbon dioxide = CO₂ (gas)
Skeleton equation: Cd(HCO₃)₂ (s) ⇒ CdCO₃ (s) + H₂O (l) + CO₂ (g)
This is already balanced.
The side reaction will affect the purity and percent yield that, it will decrease the purity and also decrease the present yield.
Due to side reaction, the amount of aspirin which is being produced will be less for some amount of starting material which is used. The present yield will decrease. The insoluble water which is being precipitated is formed as a result of the side-reaction which will remain with aspirin and finally contaminate the aspirin.
The purity of aspirin will also decrease.
Find the [OH-] in the solution. The pH is 9.5, so the pOH is 14 - 9.5 = 4.5.
[OH-] = 10^-4.5 M
Now use the dilution equation to find the new [OH-] after the volume is reduced from 150 mL to 50 mL:
M1V1 = M2V2
M1 = 10^-4.5 M
V1 = 150 mL
M2 = ?
V2 = 50 mL
(10^-4.5)(150) = M2(50)
M2 = 9.5 x 10^-5 M ≈ 1 • 10^-4 (We can only use one sig fig, because the pH was given to one decimal place.)
Now use this [OH-] to find pOH:
pOH = -log(1 x 10^-4) = 4.0
14 - pOH = pH, so the expected pH for the new solution is 10.
