A 0. 6543-g sample of that solid will require 22.39 mL of 0. 3483 m naoh for neutralization.The volume of NaOH required is 22.39 mL
The amount of NaOH needed is 22.39 mL.Sample weight is 0.6543 g.Oxalic acid mass percentage is 53.66%.This indicates that in 100 grammes of sample, 53.66 grammes of oxalic acid are present.
Oxalic acid mass in the specified sample volume = 53.66/100 ×0.6543
= 0.351g
We employ the following equation to get the number of moles:
Number of moles= Given mass/ molar
massOxalic acid mass is known to be 0.351 g.
Oxalic acid's molar mass is 90 g/mol.
Using the values in the equation above,
we obtain: 0.351/ 90 g/ mol= 0.0039 mol
The following is the chemical equation for the reaction between oxalic acid and NaOH:
C2H2O4 + 2NaOH → Na2C2O4 + 2H2O
Using the reaction's stoichiometry:
Oxalic acid reacts with two moles of NaOH in a given amount.
Therefore,
0.0039 moles of oxalic acid will react with = 2/1 × 0.0039 moles of NaOH to form a solution.
The formula used to get the molarity of the solution is used to determine the volume of the solution:
Molarity of the solution = moles of solute × 1000/ Volume of Solution (ml)NaOH equals 0.0078 moles per mole.
Solution's molarity is 0.3483 M.
Using the values in the equation above, we obtain: 0.3483 M=0.0078 × 1000/ VV = 0.0078 × 1000/ 0.3483V= 22.39 mLTherefore, 22.39 mL of NaOH are needed
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