suppose 23mL of NaOH was required. How much Carbonic Acid was neutralized?
1L of .01 NaOH contains ____ mol NaOH?
lmL of .01 NaOH contains _____mol NaOH?
23mL of .01 NaOH contains _____ mol NaOH?
2 mol of NaOH react with 1 mol H2CO3
2.3 x 10^-4 mol of NaOH react with ____mol CO2
To calculate the amount of Carbonic Acid (H2CO3) that was neutralized, we need to use the stoichiometry of the reaction and the given information.
Let's break down the solution step by step:
1. We can calculate the number of moles of NaOH in 1 liter of 0.01 M NaOH solution using the formula:
1L of 0.01 NaOH contains 0.01 mol NaOH.
2. To find the number of moles in 1 mL of 0.01 M NaOH, we divide the moles in 1 L by the total volume in liters:
1 L = 1000 mL
So, 1 mL of 0.01 NaOH contains 0.01 mol NaOH / 1000 mL = 1 x 10^-5 mol NaOH.
3. The question states that 23 mL of 0.01 M NaOH was required. To find the number of moles in 23 mL of 0.01 M NaOH, we multiply the moles in 1 mL by the volume in mL:
23 mL of 0.01 NaOH contains (1 x 10^-5 mol NaOH/mL) x 23 mL = 2.3 x 10^-4 mol NaOH.
4. It is given that 2 moles of NaOH react with 1 mole of H2CO3. So, to find the number of moles of H2CO3 neutralized, we can use the ratio:
2.3 x 10^-4 mol NaOH x (1 mol H2CO3 / 2 mol NaOH) = 1.15 x 10^-4 mol H2CO3.
Therefore, 23 mL of 0.01 M NaOH neutralizes 1.15 x 10^-4 mol of H2CO3.