What mass of a weak acid with a molar mass of 100 g/mol is necessary to neutrali
ID: 962037 • Letter: W
Question
What mass of a weak acid with a molar mass of 100 g/mol is necessary to neutralize 25 ml of 0.10 M NaOH solution?What is the pH of 0.15 g of sodium acetate NaC2H3O2 in 100 ml water H2O? What mass of a weak acid with a molar mass of 100 g/mol is necessary to neutralize 25 ml of 0.10 M NaOH solution?
What is the pH of 0.15 g of sodium acetate NaC2H3O2 in 100 ml water H2O? What mass of a weak acid with a molar mass of 100 g/mol is necessary to neutralize 25 ml of 0.10 M NaOH solution?
What mass of a weak acid with a molar mass of 100 g/mol is necessary to neutralize 25 ml of 0.10 M NaOH solution?
What mass of a weak acid with a molar mass of 100 g/mol is necessary to neutralize 25 ml of 0.10 M NaOH solution?
What is the pH of 0.15 g of sodium acetate NaC2H3O2 in 100 ml water H2O?
Explanation / Answer
a)
moles of NaOH = molarity * vol = 0.10*25/1000 = 2.5*10^-3
= moles ofweak acid = mass/mol mass
So 2.5*10^-3 = mass/100
so mass = 2.5*10^-3 * 100 = 0.25 g
b) Kb for sodium acetate is 5.65*10^-10
molar conc = 0.15(mass) /mol mass /vol in litres = 0.15/82*100/1000 = 0.0183
Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:
Ac¯ + H2O <==> HAc + OH¯
1 1 1 1
0.0183-x x x
2) Here is the Kb expression for Ac¯:
3) We can then substitute values into the Kb expression in the normal manner:
4) Ignoring the minus x in the usual manner, we proceed to solve for the hydroxide ion concentration:
x = square root of [(5.65 x 10¯10) *0.0183
x =6.78*10^-6 M = [OH¯]
We then calculate the pH. Since this is a base calculation, we need to do the pOH first:
pOH = - log 6.78*10^-6 = 5.16
pH = 14 - 5..16 = 8.8310
[HAc] [OH¯] Kb = ---------------- [Ac¯]Related Questions
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