The natural acidity of milk is 0.16% - 0.18% by mass. Predict how many mL of .1M NaOH will be required to neutralize a 50.0g sample of milk.
assuming the density of milk is near that of water, then 1 liter of milk has 1.8 grams Lactic acid
Molarity Milk= moles/liter=1.8/90=0.02M
Molaritymilk*volumilk=molaritybase*volumebase
.02*50ml=.1*Vb
Vb=.02*50/.1=10ml about
To calculate the volume of 0.1M NaOH required to neutralize a 50.0g sample of milk, you need to follow these steps:
Step 1: Determine the mass of the milk in grams.
Given in the question, the mass of the milk sample is 50.0g.
Step 2: Convert the mass of milk to percent form (0.16% to 0.18%).
The natural acidity of the milk is given as 0.16% - 0.18% by mass. We'll take the average value, which is (0.16% + 0.18%) / 2 = 0.17%.
Step 3: Convert the percent acidity to grams.
To convert the percent acidity to grams, multiply the mass of milk by the percent acidity: 0.17% × 50.0g = 0.085g.
Step 4: Calculate the number of moles of lactic acid present in the milk.
The molar mass of lactic acid (C3H6O3) is 90.08 g/mol. To calculate the number of moles, divide the mass of lactic acid by its molar mass: 0.085g / 90.08 g/mol = 0.000942 mol.
Step 5: Calculate the number of moles of NaOH required.
From the balanced chemical equation of the neutralization reaction between lactic acid (C3H6O3) and sodium hydroxide (NaOH), we know that they react in a 1:1 ratio. Therefore, the number of moles of NaOH required is also 0.000942 mol.
Step 6: Calculate the volume of 0.1M NaOH required.
Using the molarity equation, Molarity (M) = Moles (mol) / Volume (L), we can rearrange it to solve for the volume. Rearranging the equation, Volume (L) = Moles (mol) / Molarity (M), we have Volume (L) = 0.000942 mol / 0.1 mol/L = 0.00942 L.
Step 7: Convert the volume from liters to milliliters.
Multiply the volume in liters by 1000 to convert it to milliliters: 0.00942 L × 1000 = 9.42 mL.
Therefore, approximately 9.42 mL of 0.1M NaOH will be required to neutralize a 50.0g sample of milk.