Could someone please please help? Got a chem quiz tomorrow but quite confused. 1

Question

Could someone please please help? Got a chem quiz tomorrow but quite confused.

1) A 0.821 gram sample of pure NH4F was treated with 25.0 mL of 1.00M NaOH and heated to drive off the NH3.

a) How many mols of NaOH are needed to react with the NH4F?

b) How many milliliters of 0.500M HCI are needed for the back titration??

For a, I got 0.0221 moles of NH4F.

2) How is the calculated molarity of NaOH affected higher, lower, or unaffected) if the KHP used for standardization is contaminated with dirt and is not 100% pure??

Explanation / Answer

(1) Number of moles of NH4F , n = mass/molar mass

= 0.821 g / 37 (g/mol)

= 0.022 moles

Number of moles of NaOH , n' = Molarity x volume in L

= 1.00 M x 25.0 mL x 10-3 L/mL

= 0.025 mol

The balanced reaction is : NH4F + NaOH ----> NH3 + NaF + H2O

1 mole of NH4F reacts with 1 mole of NaOH

0.022 moles of NH4F reacts with 0.022 moles of NaOH -----(a)

So (0.025 - 0.022) = 0.003 moles of NaOH left unreacted so NaOH is the excess reactant

Since all of NH4F completly reacted NH4F is the limiting reactant

NaOH + HCl ---> NaCl + H2O

1 mole of NaOH reacts with 1 mole of HCl

0.003 moles of NaOH reacts with 0.003 moles of HCl

Therefore The volume of HCl needed , V (in L) = number of moles / Molarity

= 0.003 mol / 0.500 M

= 0.006 L

= 6 mL ------> (b)

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