pH of the buffer solution is 1.76.
Chemical dissociation of formic acid in the water:
HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)
The solution of formic acid and formate ions is a buffer.
[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions
[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate
[HCOOH] = 1.45 M - 0.015 M
[HCOOH] = 1.435 M; equilibrium concentration of formic acid
pKa = -logKa
pKa = -log 1.8×10⁻⁴ M
pKa = 3.74
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)
pH = 3.74 + log (0.015 M/1.435 M)
pH = 3.74 - 1.98
pH = 1.76
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Answer:
No.of moles of C is , n = mass/molar mass = 75.46 g / 12 (g/mol) = 6.3 moles No.of moles of H is , n' = mass/molar mass = 4.43 g / 1.0(g/mol) = 4.43 moles No.of moles of O is , n'' = mass/molar mass = 20.10 g / 16(g/mol) =1.25 moles Ratio to the no.of moles of C,H& O is 6.3 : 4.43 : 1.25 In the simple integer ratio is ( 6.3/1.25) : ( 4.43/1.25) : (1.25/1.25) 5.04 :3.5 : 1
Explanation:
Answer: 2.8 moles
Explanation:
The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.
Na2O + H2O --> 2NaOH
1 mole of H2O = 2 moles of NaOH
Let Z moles of H2O = 5.6 mole of NaOH
To get the value of Z, cross multiply
5.6 moles x 1 mole= Z x 2 moles
5.6 = 2Z
Divide both sides by 2
5.6/2 = 2Z/2
2.8 = Z
Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH
The ph would most likely be 7
hope this helps
