CH₃COOH millimoles = 52 x 0.35 = 18.2
NaOH millimoles = 27 x 0.4 = 10.8
CH₃COOH + NaOH → CH₃COONa + H₂O
18.2 10.8 0 0 -----> Initial
- 10.8 - 10.8 + 10.8 + 10.8 -----> Changed
7.4 0 10.8 10.8 ------> after reaction
Now the mixture contain acid and salt so it forms buffer
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
pH = pKa + log [salt] / [acid]
pH = 4.74 + log [10.8] / [7.4]
pH = 4.9
Can you show me the passage?
Answer: (The composition of a bituminous coal by percentage is roughly: carbon [C], 75–90; hydrogen [H], 4.5–5.5; nitrogen [N], 1–1.5; sulfur [S], 1–2; oxygen[O], 5–20; ash, 2–10; and moisture, 1–10.)
Answer:
A. 2Fe + 3Cl2 --> 2FeCl3
B. 2Fe + 3O2 --> 2Fe2O3
C. C4H10O + 4O2 --> 4CO2 + 5H2O
D. C7H16 + 11O2 --> 7CO2 + 8H2O
Explanation:
3 to balance the calciums on both sides of the equation, but only if that balances the Cl on both sides of the equation to because you now have 6 Cl and 3Ca