If I spill 100ml of a .5 M H3PO4 solution, how many grams of NaOH should be dissolved in water and added to the spill to neutralize the acid?
H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
mols H3PO4 = M x L = ?
mols NaOH = 3x that.
mols NaOH = grams/molar mass. You know mols and molar mass, solve for grams.
To answer this question, we need to use the concept of stoichiometry. Stoichiometry is a way to calculate the amount of substances involved in a chemical reaction, based on their balanced equation.
First, let's balance the equation for the neutralization reaction between H3PO4 and NaOH:
H3PO4 + 3NaOH → Na3PO4 + 3H2O
From the balanced equation, we can see that one molecule of H3PO4 reacts with three molecules of NaOH. Therefore, the stoichiometric ratio between H3PO4 and NaOH is 1:3.
To calculate how many moles of H3PO4 you spilled, we need to use the formula:
moles = concentration (M) × volume (L)
Given that the concentration of the H3PO4 solution is 0.5 M and the volume spilled is 100 mL (which is equivalent to 0.1 L), we can calculate the number of moles of H3PO4:
moles of H3PO4 = 0.5 M × 0.1 L = 0.05 moles
Since the stoichiometric ratio between H3PO4 and NaOH is 1:3, we need three times as many moles of NaOH to neutralize the acid. Therefore, we need:
moles of NaOH = 3 × moles of H3PO4 = 3 × 0.05 moles = 0.15 moles
Now we need to convert the moles of NaOH to grams. To do this, we need to know the molar mass of NaOH, which is 22.99 g/mol for sodium (Na), 16.00 g/mol for oxygen (O), and 1.01 g/mol for hydrogen (H). Adding these together, we get:
molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40 g/mol
Finally, we can calculate the grams of NaOH needed:
grams of NaOH = moles of NaOH × molar mass of NaOH = 0.15 moles × 40 g/mol = 6 grams
Therefore, you should dissolve 6 grams of NaOH in water and add it to the spill to neutralize the H3PO4 solution.