Question

If possible please give the formula or steps required for me to solve these questions, as I would like to go through the steps myself rather than just recieving answers. Thanks.

PART 1

Mass of Weighing Paper: 0.3966g

Mass of Weighing Paper & Oxalic Acid: 0.4705g

Initial Buret Reading: 0.00mL

Final Buret Reading: 12.80mL

1.) Moles of Oxalic Acid used?

2.) Protons (H+) avalivable for reaction with hydroxide ion (OH-)?

3.) Moles of hydroxide ion (OH-) which reacted with protons (H+)?

4.) Precise Molarity of NaOH solution?

PART 2

Proposed mass of unknown acid: 0.189g

Number of reactive protons per molecule: 2

Mass of weighing paper: 0.3958g

Mass of weighing paper and unknown acid: 0.4751g

Initial Buret Reading: 0.00mL

Final Buret Reading: 10.80mL

1.) Moles of NaOH which reacted with unknown acid?

2.) Moles of unknown acid which reacted with hydroxide ion (OH-)?

3.) Molecular weight of your unknown acid?

Answer 1

ANSWER - Part I:

a) Moles of oxalic acid used?

• To calculate the moles of oxalic acid we use this formula:

$moles=\frac{mass}{Molecular\, Weight}$

• The molecular weight of oxalic acid (C₂H₂O₄) is 90.03 g/mol
• The mass of oxalic acid used is

$mass\, C_{2}H_{2}O_{4}=(Weighing\, paper\, \&\, oxalic\, acid)-(Weighing\, paper)$

$mass\, C_{2}H_{2}O_{4}=(0.4705\, g)-(0.3966\, g)=\mathbf{0.0739\, g}$

• Then, the moles of oxalic acid used are

$moles=\frac{0.0739\, g}{90.03\, g/mol}=\mathbf{8.21x10^{-4}\, mol}$

b) Protons (H⁺) available for reaction with hydroxyde ion (OH^-)?

• The oxalic acid (C₂H₂O₄) ihas the following structure:.

ОН Но-

• It is a dicarboxilic acid, this means that each molecule of oxalic acid has two protons available to react with OH^-. Then, 1 mol of oxalic acid has 2 moles of protons:

$1\, mol\, C_{2}H_{2}O_{4}\Rightarrow 2\, mol\, H^{+}$

$8.21x10^{-4}\, mol\, C_{2}H_{2}O_{4}\Rightarrow \mathbf{1.64x10^{-3}\, mol\, H^{+}}$

c) Moles of hydroxide ion (OH^-) which reacted with protons (H⁺)?

• A neutralzation reaction has the following equation:

$H^{+}+OH^{-}\rightarrow H_{2}O$

• This means that 1 mol of protons reacts with 1 mol of hydroxyde ion. Then, the moles of OH^- used are:

$1\, mol\, H^{+}\Rightarrow 1\, mol\, OH^{-}$

$1.64x10^{-3}\, mol\, H^{+}\Rightarrow \mathbf{1.64x10^{-3}\, mol\, OH^{-}}$

d) Precise molarity of NaOH solution?

• The molarity of solution is defined as

$Molarity=\frac{moles\, solute}{Volume\, solution(L)}$

• The moles of NaOH are

$1\, mol\, OH^{-}\Rightarrow 1\, mol\, NaOH$

$1.64x10^{-3}\, mol\, OH^{-}\Rightarrow \mathbf{1.64x10^{-3}\, mol\, NaOH}$

• The volume of NaOH solution is

$V_{NaOH}=Final\, reading-Initial\, reading$

$V_{NaOH}=12.80\, mL-0.00\, mL=12.80\, mL$

$V_{NaOH}=\mathbf{12.80x10^{-3}\, L}$

• Then, the molarity NaOH solution is

$Molarity=\frac{1.64x10^{-3}\, mol}{12.80x10^{-3}\, L}=\mathbf{0.128\, M}$

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ANSWER - Part II:

a) Moles of hydroxide ion (OH^-) which reacted with unknown acid?

• We know the volume of 0.128 M NaOH solution used for reaction with unknown acid:

$V_{NaOH}=Final\, reading-Initial\, reading$

$V_{NaOH}=10.80\, mL-0.00\, mL=10.80\, mL$

$V_{NaOH}=\mathbf{10.80x10^{-3}\, L}$

• Now, we can calculate the moles of NaOH used

$moles\, NaOH=M_{NaOH}\times V_{NaOH}$

$moles\, NaOH=0.128\, \frac{mol}{L} \times 10.80x10^{-3}L=\mathbf{1.38x10^{-3}\, mol}$

b) Moles of unknown acid which reacted with hydroxide ion (OH^-)?

• Each mol of OH^- (it comes from NaOH) reacts with one mol of H⁺ (it comes from unknown acid)

$H^{+}+OH^{-}\rightarrow H_{2}O$

• Then:

$1\, mol\, NaOH=1\, mol\, OH^{-}\Rightarrow 1\, mol\, H^{+}$

$1.38x10^{-3}\, mol\, OH^{-}\Rightarrow \mathbf{1.38x10^{-3}\, mol\, H^{+}}$

• In addition, one molecule of unknown acid has 2 reactive protons, then

$2\, mol\, H^{+}\Rightarrow 1\, mol\, unknown\, acid$

$1.38x10^{-3}\, mol\, H^{+} \Rightarrow \mathbf{6.9x10^{-4}\, mol\, unknown\, acid}$

c) Molecular weight of your unknown acid?

• The molecular weight of unknown acid is defined as

$Molecular\, Weight=\frac{mass}{moles}$

• The mass of unknown acid used is

$mass\, unknown\, acid=(Weighing\, paper\, \&\, unknown\, acid)-(Weighing\, paper)$

$mass\, unknown\, acid=(0.4751\, g)-(0.3958\, g)=\mathbf{0.0793\, g}$

• Then,

$Molecular\, Weight=\frac{0.0793\, g}{6.9x10^{-4}\, mol}=\mathbf{114.93\, g/mol}$​​​​​​​