Titrimetric analysis?
a. What is the molar concentration of the NaOH solution?
b. Calculate the gramsof NaOH in the solution
Answer:
The balanced equation is
NaOH + HCl >> NaCl + H2O
Therefore 1 mol of HCl reacts with
1mol NaOH stoichiometrically
Therefore
0.2340 moles HCl x 0.01740L = 4.072E-3 Mol HCl
...1L HCl
Considering there is a 1:1 mole ratio you have
4.072E-3 moles of NaOH
4.072Mol NaOH = 2.7E-1 M NaOH
..0.015L
Now for the number of grams:
4.072E-3 moles of NaOHx(39.9971gNaOH/1molNaOH)=
1.629E-1 grams of NaOH
The reaction can be written as:
NaOH + HCl ----> NaCl + H2O
So, we have one mole of NaOH reacting for each mole of HCl added. So, the number of moles of HCl reacted = 0.0174 * 0.234 = 4.0716 * 10^-3 moles
So, the number of moles of NaOH reacted = 4.0176*10^-3 moles
15 mL of the sample contains 4.0176*10^-3 moles.
a)
So, the molarity (number of moles in 1000 mL) = 4.0176*10^-3*1000/15 = 0.271 M
b)
The number of moles of NaOH in the solution = 4.0176*10^-3 moles
So, the number of grams of NaOH = Molecular wt * Number of moles = 40*4.0176 *10^-3=0.1607 grams
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