A student dilutes 50.0 mL of a 0.10 mol/L to 0.010 mol/L. Which statement is true?
1 answer:
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Answer:
pH of the final solution = 3.8
Explanation:
Concentration of NaF = molar
= 0.3 molar
NaF → Na⁺ + F⁻
HF ⇆ H⁺ + F⁻
- NaF is strong electrolyte so completely ionized but HF weak acid not completely ionized.
- Since F⁻ is common ion here
according to common ion effect dissociation of weak acid decreases.
Ka =
⇒ [H⁺] = ...............(1)
{Ka of HF = 3.5 x 10⁻⁴} & Concentration of HF = 30 x 4 x 10⁻³ = 0.12 molar
from equation 1
⇒ [H⁺] = [Concentration of F⁻ ≡ Concentration of NaF]
⇒ [H⁺] = 0.00014
⇒pH = - log 0.00014 = 3.85
Answer:
The amount of NaOH required to prepare a solution of 2.5N NaOH.
The molecular mass of NaOH is 40.0g/mol.
Explanation:
Since,
NaOH has only one replaceable -OH group.
So, its acidity is one.
Hence,
The molecular mass of NaOH =its equivalent mass
Normality formula can be written as:
Substitute the given values in this formula to get the mass of NaOH required.
Hence, the mass of NaOH required to prepare 2.5N and 1L. solution is 100g