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 %
Hey there!
To calculate the percent by mass of the Ca(NO₃)₂ we need to find the total mass first by adding.
896.92 + 22.63 = 919.55
In total, the solution is 919.55 grams.
To find the percent of Ca(NO₃)₂ in the solution, divide the mass of Ca(NO₃)₂ by the total mass and multiply by 100.
22.63 ÷ 919.55 = 0.0246
0.0246 x 100 = 2.46
Ca(NO₃)₂ makes up 2.46% of the solution.
Hope this helps!
11. ionic charge +1, helium.
12. ionic charge 2-, neon.
13. ionic charge 3+, neon.
Answer:
![Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OH%5D%7D)
Explanation:
Hello,
In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:
Thus, we can write the law of mass action, which consider the equilibrium concentrations of all the involved species, which is also known as the acid dissociation constant which accounts for the capacity the acid has to yield hydronium ions:
![K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}](https://tex.z-dn.net/?f=K%3DKa%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OH%5D%7D)
Best regards.
