2. Suppose you measured out 0.20 g of tartaric acid, a diprotic acid, H2C4H4O6 , determine the volume of 0.15 M sodium hydroxide required to neutralize this acid.
Neutralization equation:
Moles acid= 2*moles Base (balance the equation to see the 2)
.2/150=2*.15*Vb(in liters)
Vb=.2/(150*2*.15) =.004 liters check that
To determine the volume of 0.15 M sodium hydroxide required to neutralize 0.20 g of tartaric acid (H2C4H4O6), you need to follow the steps below:
Step 1: Calculate the number of moles of tartaric acid.
To do this, you need to know the molar mass of tartaric acid. The molar mass of H2C4H4O6 is:
2(1.01 g/mol) + 4(12.01 g/mol) + 6(1.01 g/mol) + 6(16.00 g/mol) = 150.09 g/mol
Now, you can calculate the number of moles of tartaric acid:
moles of tartaric acid = mass of tartaric acid / molar mass of tartaric acid
moles of tartaric acid = 0.20 g / 150.09 g/mol
Step 2: Determine the stoichiometry between tartaric acid and sodium hydroxide.
From the equation: H2C4H4O6 + 2NaOH -> 2H2O + Na2C4H4O6
We can see that 1 mole of tartaric acid reacts with 2 moles of sodium hydroxide.
Step 3: Calculate the volume of sodium hydroxide required.
The molarity (M) of sodium hydroxide is given as 0.15 M.
Using the equation:
moles of sodium hydroxide = moles of tartaric acid * (2 moles of sodium hydroxide / 1 mole of tartaric acid)
Now, we can calculate the moles of sodium hydroxide:
moles of sodium hydroxide = (0.20 g / 150.09 g/mol) * (2 mol of NaOH / 1 mol of H2C4H4O6)
Step 4: Calculate the volume of sodium hydroxide using its molarity.
Using the equation:
volume of sodium hydroxide = moles of sodium hydroxide / molarity of sodium hydroxide
Now, we can calculate the volume of sodium hydroxide:
volume of sodium hydroxide = moles of sodium hydroxide / 0.15 mol/L
By substituting the values and calculating, you will determine the volume of 0.15 M sodium hydroxide required to neutralize this acid.
To determine the volume of 0.15 M sodium hydroxide required to neutralize 0.20 g of tartaric acid, we need to follow several steps.
1. Calculate the number of moles of tartaric acid:
- First, we need to find the molar mass of tartaric acid. The molar mass of H2C4H4O6 can be calculated by adding the atomic masses of its elements:
- Hydrogen (H) has a molar mass of 1 g/mol, Carbon (C) has a molar mass of 12 g/mol, and Oxygen (O) has a molar mass of 16 g/mol.
- The molar mass of tartaric acid (H2C4H4O6) = (2 x 1 g/mol) + (4 x 12 g/mol) + (6 x 16 g/mol) = 150 g/mol.
- To find the number of moles of tartaric acid, we divide its mass by its molar mass:
- Moles of tartaric acid = Mass of tartaric acid / Molar mass of tartaric acid
- Moles of tartaric acid = 0.20 g / 150 g/mol = 0.00133 mol
2. Use the stoichiometry of the reaction to find the number of moles of sodium hydroxide (NaOH) required to neutralize tartaric acid.
- The balanced equation for the neutralization reaction between tartaric acid and sodium hydroxide is:
- H2C4H4O6 + 2NaOH → Na2C4H4O6 + 2H2O
- From the equation, we can see that 1 mole of tartaric acid reacts with 2 moles of sodium hydroxide.
- Moles of sodium hydroxide required = Moles of tartaric acid x (2 moles of NaOH / 1 mole of H2C4H4O6)
- Moles of sodium hydroxide required = 0.00133 mol x (2/1) = 0.00266 mol
3. Finally, calculate the volume of 0.15 M sodium hydroxide required.
- We can use the formula: Volume (V) = Moles (n) / Concentration (C)
- Volume of sodium hydroxide required = Moles of sodium hydroxide required / Concentration of sodium hydroxide
- Volume of sodium hydroxide required = 0.00266 mol / 0.15 mol/L = 0.0177 L or 17.7 mL
Therefore, you would need approximately 17.7 mL (or 0.0177 liters) of 0.15 M sodium hydroxide to neutralize 0.20 g of tartaric acid.
This is a diprotic weak acid with a pKa₁ of 2.89 => Ka₁ = 1.3 x 10⁻³ and a pKa₂ = 4.40 => Ka₂ = 4.0 x 10⁻⁵. With such a low Ka₂, it is assumed (generally) that all of the Hydronium ions come from the 1st Ionization Step.
For 0.20-g H₂Tartarate (=0.2/150 mole) => 0.0013-mole Tartaric Acid being added into solution.
Assuming a 1-Liter solution => 0.0013M Tartartic Acid (Ka₁ = 1.3 x 10⁻³) delivers ~ 0.0013M H₃O⁺ in solution.
Assuming all Hydronium ions are from the 1st ionization step & using the ICE Table analysis => [H₃O⁺] = SqrRt(Ka∙[Tartaric Acid]) = SqrRt((1.3 x 10⁻³)²) = 0.0013M H₃O⁺ => 0.0013-mole/Liter, needs neutralizing.
Molarity x Volume = moles => Volume(L) = moles/Molarity = 0.0013-mole/0.15-M = 0.0087-L = 8.7-ml 0.15M NaOH(aq).