If 25.0 cm3 of 0.50M sulfuric acid reacted with exactly 12.5 cm3 of sodium hydroxide, calculate the concentration of the alkali.
I first balanced the chemical equation between sulfuric acid and sodium hydroxide.
H2SO4 + 2NaOH --> 2H2O + Na2SO4
I converted the volume of sulfuric acid to litres then to mols. I multiplied by the mol ratio to get the moles of NaOH which was .025 mols then divided by the volume of the base to get 2M NaOH. Is this correct?
yes
Yes, you have taken the correct steps to find the concentration of the sodium hydroxide (NaOH) solution. Let's go through the calculations to confirm if your approach is correct.
To find the concentration of the alkali (NaOH), you first balanced the chemical equation between sulfuric acid (H2SO4) and sodium hydroxide (NaOH), which is:
H2SO4 + 2NaOH --> 2H2O + Na2SO4
Next, you converted the volume of sulfuric acid (25.0 cm3) to liters (L). Since 1 cm3 is equal to 1 mL, you have:
25.0 cm3 = 25.0 mL = 0.025 L
Then, you used the molarity (M) formula, which is moles of solute divided by liters of solution, to find the moles of sulfuric acid. Since the sulfuric acid concentration is given as 0.50 M, the moles can be calculated as:
0.50 mol/L * 0.025 L = 0.0125 mol H2SO4
Now, referring to the balanced equation, you can see that the mole ratio between sulfuric acid and sodium hydroxide is 1:2. Therefore, for each mole of sulfuric acid, you will need 2 moles of sodium hydroxide.
Since you have 0.0125 moles of sulfuric acid, you can calculate the moles of sodium hydroxide by multiplying by the mole ratio:
0.0125 mol H2SO4 * (2 mol NaOH / 1 mol H2SO4) = 0.025 mol NaOH
Finally, to obtain the concentration of the sodium hydroxide solution, divide the moles of NaOH by the volume of sodium hydroxide (12.5 cm3 converted to L):
0.025 mol / 0.0125 L = 2 M
So, based on your calculations, the concentration of the alkali (sodium hydroxide) is indeed 2 M. Well done!