Answer:
3.02
Explanation:
Step 1: Given and required acid
pH of the solution: 4.34
pKa of lactic acid: 3.86
Step 2: Calculate the ratio of the lactate concentration to the lactic acid concentration in this solution
We have a buffer system formed by a weak acid (lactic acid) and its conjugate base (lactate ion). We can find the ratio of the lactate concentration to the lactic acid concentration in this solution using the Henderson-Hasselbach equation.
pH = pKa + log [lactate]/[lactic acid]
4.34 = 3.86 + log [lactate]/[lactic acid]
[lactate]/[lactic acid] = 3.02
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
Total moles of gas = 0.1225
Volume of gas produced : 6.7375 L
mass of Nitrogen : 0.588 g
<h3>Further explanation</h3>
Given
2 ml of Nitroglycerin(ρ=1.592 g/ml)
Required
Total moles of gas
Solution
Nitroglycerin detonated ⇒ decomposition reaction
4C₃H₅N₃O₉(s)⇒ 6N₂(g)+12CO(g)+10H₂O(g)+7O₂(g)
mass of Nitroglycerin :
moles of Nitroglycerin :
Total moles of gas:
Volume of gas produced :
moles of Nitrogen :
mass of Nitrogen :