Answer:
O.1M
Explanation:
First let's generate a balanced equation for the reaction
NaOH + HCl —>NaCl + H2O
From the equation,
The ratio of the acid to base is 1:1.
From the question, we obtained the following:
Ma = Molarity of acid = 0.12M
Va = volume of acid = 21.35cm3
Vb = volume of base = 25.55cm3
Mb = Molarity of base =?
We obtained nA(mole of acid) and nB(mole of base) to be 1
The molarity of the base can be calculated for using:
MaVa/ MbVb = nA / nB
0.12x21.35 / Mb x 25.55 = 1
Cross multiply to express in linear form
Mb x 25.55 = 0.12x21.35
Divide both side by 25.55
Mb = (0.12x21.35) / 25.55
Mb = 0.1M
The molarity of the base is 0.1M
Hey there!:
Number of moles = ( number of atoms / 6.023*10²³ atoms )
given number of atoms = 5.03*10²⁴
Therefore:
Number of moles B = 5.03*10²⁴ / 6.023*10²³
Number of moles B = 8.35 moles
Hope that helps!
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 %
