1) Why is it necessary to standardize an NaQH solution? 2) Write the balanced ch
ID: 593091 • Letter: 1
Question
1) Why is it necessary to standardize an NaQH solution? 2) Write the balanced chemical equation for the reaction of NaOH with HCl. ) Look up the molecular weight for potassium hydrogen phthalate (KHP) and determine the number moles of KHP in 1.85 g of KHP 4An NaOH solution of unknown concentration is used to titrate 1.85 g of KHP. The initial reading of the burette is: M8.35 ml, The student titrates the sample to the point where the solution has just turned pink. The final burette reading gives V19.40 m Calculate the concentration of the NaQH solution. 5) If the NaOH solution from question 4 is used to titrate 25 mL of 0.682 M HCl, how much NaOH will be required?Explanation / Answer
Ans. #1. Both solid NaOH and its solution is hydroscopic, i.e. absorb moisture from atmosphere. So, mass of NaOH pellets may usually be contaminated with moisture and using mass to calculate molarity may yield errors. Similarly, the concertation of an standardized NaOH solution may also decrease slightly when left for some time (say, few days or weeks).
Therefore, standardization of NaOH after preparing it and after re-using it required to know the exact molarity of NaOH solution at the time of its use.
#2. Balanced Reaction: NaOH(aq) + HCl(aq) ----------> NaCl(aq) + H2O(l)
#3. Molar mass of KHP = 204.22 g/mol
Now,
Moles of KHP = Mass / Molar mass
= 1.85 g / (204.22 g/ mol)
= 0.009059 mol
#4. Volume of NaOH consumed = Final reading – Initial reading
= 19.40 mL – 8.35 mL
= 11.05 mL
= 0.01105 L
# Balanced Reaction: KHC8H4O4(aq) + NaOH(aq) -------> KNaC8H4O4(aq) + H2O(l)
According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol KHP.
KHC8H4O4(aq) is better known as KHP.
So,
Moles of NaOH consumed = Moles of KHP in 1.85 g sample
Hence, Moles of NaOH consumed = 0.009059 mol
Now,
Molarity of NaOH solution = Moles of NaOH / Volume of solution in liters
= 0.009059 mol / 0.01105 L
= 0.81982 M
Hence, [NaOH] = 0.81982 M
#5. Using C1V1 (NaOH) = C2V2 (HCl)
Or, 0.81982 M x V1 = 0.682 x 25.0 mL
Or, V1 = (0.682 x 25.0 mL) / 0.81982
Hence, V1 = 20.797 mL
Hence, required volume of NaOH solution = 20.797 mL
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