The formula weight of an acid is 82. In a titration, 100cm3 of the solution of this acid containing 39g of this acid per litre were completely neutralized by 95 cm3 of aqueous naoh containing 40g of naoh per litre. What is the basicity of this acid?
Bob helps me
To find the basicity of the acid, we need to determine the equivalent weight of the acid and the number of moles of acid that reacted with the NaOH.
Step 1: Calculate the equivalent weight of the acid.
The formula weight of the acid is given as 82g/mol. The equivalent weight is the weight of the acid that corresponds to one mole of H+ ions. Since the acid is monoprotic (donates only one H+ ion), the equivalent weight is equal to the formula weight.
Equivalent weight of the acid = 82g/mol
Step 2: Calculate the number of moles of acid that reacted with NaOH.
From the titration, we know that the volume of the acid solution that reacted is 100 cm3 or 0.1 L, and its concentration is given as 39g/L.
Number of moles of acid = (Volume of acid solution x Concentration of acid) / Formula weight
= (0.1 L x 39g/L) / 82g/mol
Step 3: Calculate the basicity of the acid.
The basicity of an acid is the number of moles of H+ ions it can donate to neutralize a base. In this case, the acid reacts with NaOH, which is a base that donates one mole of OH- ions.
Basicity of the acid = Number of moles of acid / Number of moles of NaOH
Now, we need to determine the number of moles of NaOH that reacted. Given that the volume of NaOH solution used is 95 cm3 or 0.095 L, and its concentration is given as 40g/L.
Number of moles of NaOH = (Volume of NaOH solution x Concentration of NaOH) / Formula weight of NaOH
= (0.095 L x 40g/L) / 40g/mol
= 0.095 mol
Substituting this value into the equation:
Basicity of the acid = (0.1 mol) / (0.095 mol)
Therefore, the basicity of this acid is approximately 1.053.
To find the basicity of the acid, we need to understand the concept of basicity and how it relates to the titration process.
Basicity refers to the number of replaceable hydrogen atoms in an acid molecule. It is determined by the number of hydroxide ions (OH-) required to neutralize one mole of the acid. This can be found by analyzing the balanced chemical equation for the reaction between the acid and base.
In this question, we are given information about the acid and the base used in the titration.
The formula weight of the acid is given as 82, which means that one mole of the acid weighs 82 grams.
We are also given that 100 cm3 of the acid solution containing 39 g of the acid per liter were neutralized by 95 cm3 of aqueous NaOH containing 40 g of NaOH per liter.
To solve this problem, we need to determine the moles of acid and base used in the titration.
Moles of acid = mass of acid / molar mass of acid
Moles of acid = 39 g / 82 g/mol = 0.4762 mol
Moles of base = volume of NaOH x concentration of NaOH
Moles of base = 0.095 L x (40 g/L / 40 g/mol) = 0.095 mol
From the balanced chemical equation, we know that one mole of acid reacts with one mole of base in a 1:1 ratio.
Since the moles of acid and base used in the titration are equal (0.4762 mol = 0.095 mol), we can conclude that the basicity of the acid is 1.
Therefore, the basicity of this acid is 1.
mols NaOH in solution = grams/formula weight = 40/40 = 1 mol.
M NaOH = 1 mol/L = 1M
mols acid = grams/formula weight = 39/82 = about 0.48 but you do it more accurately.
M base = mols/L = 0.48M
mols acid x #H = mols base
0.48M x 0.100L x #H = 1M x 0.095L
#H = (1M x 0.095)/(0.48 x 0.095)
#H = ?