What is the concentration (m) of sodium ions in 4.57 l of a 2.35 m na3p solution?
2 answers:
Answer:
7.05 M
Explanation:
- An equation can be written:
Na₃P → 3Na⁺ + P⁻³
- The formula for concentration (M) is the number of moles divided by the volume in L:
C = n / V
With the concentration and volume given by the problem, we know the number of moles of Na₃P, then with that number we can calculate the moles of Na⁺:
4.57 L * 2.35 M = 10.7395 mol Na₃P
10.7395 mol Na₃P * = 32.2185 mol Na⁺.
Concentration of Na⁺ = mol Na⁺ / V
[Na⁺] = 32.2185 mol Na⁺ / 4.57 L = 7.05 M
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Answer:
The pH is 7.54
Explanation:
The Henderson - Hasselbalch equation states that for a buffer solution which consists of a weak acid and its conjugate base, the buffer pH is given by:
pH
pkₐ is for the acid
In this case, the buffer hypochlorous acid HClO is a weak acid, and its conjugate base is the hypochlorite anion ClO⁻ is delivered to the solution via sodium hypochlorite NaClO .
NaCIO = 0.200 M
HCIO = 0.200 M
pkₐ = -log₁₀ kₐ = -log₁₀ (2.9 × 10⁻⁸) = 7.54
∴pH = = 7.54