What is the (H30) concentration when 40 mL of 0.10 M nitrous acis, HNO2, are added to 10.0 mL of 0.10 M HCl? Ka for HNO2 is 4.0 x 10^-4.
I get 1 x 10^-3. Is this right?
I didn't get your answer.
millimoles HCl = 10.0 x 0.1 = 1 mmole.
(HCl) = 1 mmole/50 mL = 0.02 M
HNO2 ==> H^+ + NO2^-
The H^+ from HCl is a common ion here; therefore, the equilibrium for HNO2 is shifted to the left.
Ka = (H^+)(NO2^-)/(HNO2)
Substitute 0.02 for (H^+). Calculate (HNO2) = 40 x 0.1 = 4 millimoles and that divided by 50 mL = 0.08 M. Then add H^+ from HCl and NO2^- (that will be the equivalent H^+ from the HNO2) to obtain total H^+. Something like 0.022 or close to that.
To determine the (H3O+) concentration, we can use the concept of stoichiometry and acid-base equilibrium. Here's how you can calculate it:
First, let's write the balanced chemical equation for the reaction between HNO2 and HCl:
HNO2 + HCl --> H3O+ + NO2-
Since the reaction is 1:1, we know that the moles of H3O+ produced will be equal to the moles of HNO2 consumed.
Next, calculate the moles of HNO2 and HCl. To do this, we need to use the formula:
moles = concentration (mol/L) × volume (L)
For HNO2:
moles of HNO2 = 0.10 mol/L × 0.040 L = 0.004 mol
For HCl:
moles of HCl = 0.10 mol/L × 0.010 L = 0.001 mol
Based on the balanced equation, we see that 1 mol of HNO2 reacts to produce 1 mol of H3O+. Therefore, the moles of H3O+ produced will also be 0.004 mol.
Now, we need to calculate the concentration of (H3O+). To do this, divide the moles of H3O+ by the total volume of the solution.
Total volume = 40 mL + 10 mL = 50 mL = 0.050 L
(H3O+) concentration = moles of H3O+ / total volume = 0.004 mol / 0.050 L = 0.08 M
So, the correct answer is (H3O+) concentration = 0.08 M.
Therefore, your initial answer of 1 x 10^-3 is incorrect.