How many grams of glycolic acid are needed to react with 30 mL of 0.1 M NaOH?
CH2OHCOOH + NaOH ==> CH2OHCOONa + H2O
mols NaOH = M x L = ?
Using the coefficients in the balanced equation, convert mols NaOH to mols glycolic acid. That should be 1:1.
Then mols glycolic acid = grams/molar mass. You know molar mass and mols, solve for grams.
To determine the number of grams of glycolic acid needed to react with 30 mL of 0.1 M NaOH, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction between glycolic acid (HA) and sodium hydroxide (NaOH).
HA + NaOH → NaA + H2O
Step 2: Determine the molar ratio between glycolic acid and sodium hydroxide from the balanced equation.
From the balanced equation, we can see that the molar ratio between glycolic acid and sodium hydroxide is 1:1. This means that, for every 1 mole of NaOH, we need 1 mole of glycolic acid.
Step 3: Calculate the number of moles of NaOH in 30 mL of 0.1 M NaOH.
To do this, we'll use the formula:
moles = concentration (M) x volume (L)
Given that the concentration of NaOH is 0.1 M and the volume is 30 mL (which is 0.03 L after converting from mL to L), we can use the formula:
moles of NaOH = 0.1 M x 0.03 L = 0.003 moles NaOH
Step 4: Calculate the number of moles of glycolic acid needed.
Since the molar ratio between glycolic acid and NaOH is 1:1, the number of moles of glycolic acid needed is also 0.003 moles.
Step 5: Determine the molar mass of glycolic acid.
The molar mass of glycolic acid (C2H4O3) can be calculated by adding up the atomic masses of its constituent elements. The atomic masses of carbon (C), hydrogen (H), and oxygen (O) are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively.
Molar mass of C2H4O3 = (2 x atomic mass of C) + (4 x atomic mass of H) + (3 x atomic mass of O)
= (2 x 12.01 g/mol) + (4 x 1.01 g/mol) + (3 x 16.00 g/mol)
= 24.02 g/mol + 4.04 g/mol + 48.00 g/mol
= 76.06 g/mol
Step 6: Calculate the mass of glycolic acid needed.
To calculate the mass of glycolic acid needed, we'll use the formula:
mass = moles x molar mass
mass of glycolic acid = 0.003 moles x 76.06 g/mol
= 0.22818 g
Therefore, approximately 0.228 grams of glycolic acid are needed to react with 30 mL of 0.1 M NaOH.