My final exam is coming up so I am doing study guides to help me practice. I would really appreciate any help that is given.
1. The hydronium ion concentration for a solution is determined to be
4.5 x 10^-7 M. What is the concentration of the hydroxide ion?
2. When a 5.00 mL sample of vinegar is titrated, 44.5 mL of 0.100 M NaOH solution is required to reach the end point. What is the concentration of vinegar in moles per liter of solution?
I'm not really sure what numbers to put where so if you could please walk me through how to do this so I can use it as a how-to for other problems like it on the study guide I would greatly appreciate it.
1.
(H^+)(OH^-) = Kw = 1E-14
Substitute for H^+ and solve for OH^-
2.
vinegar is acetic acid, HAc.
HAc + NaOH ==> NaAc + H2O
mols NaOH = M x L = ?
mols HAc = mols NaOH (Look at the coefficients in the balanced equation.) That's mols in 5.00 mL of the solution.
Then M HAc = mols HAc/L soln.
(you know L soln = 0.005L)
Thank you. That helps.
Of course! I'm here to help you through these problems step by step.
1. To determine the concentration of the hydroxide ion, we can use the concept of the ion product of water (Kw). The ion product of water is the product of the hydronium ion (H3O+) concentration and the hydroxide ion (OH-) concentration, and is equal to 1.0 x 10^-14 at 298 K.
Since we know the hydronium ion concentration (H3O+) is 4.5 x 10^-7 M, we can use this information to find the concentration of the hydroxide ion (OH-). Let's call the concentration of the hydroxide ion x.
Using the ion product of water equation, Kw = [H3O+][OH-], we can substitute in the given values:
1.0 x 10^-14 = (4.5 x 10^-7)(x)
Now, solve for x by rearranging the equation:
x = (1.0 x 10^-14) / (4.5 x 10^-7)
To simplify this, divide the numerator and denominator by 4.5 x 10^-7:
x = (1.0 x 10^-14) / (4.5 x 10^-7) = 2.2 x 10^-8
Therefore, the concentration of the hydroxide ion (OH-) is 2.2 x 10^-8 M.
2. In this problem, we're given the volume and concentration of the NaOH solution used for titration, as well as the volume of vinegar used. We need to calculate the concentration of vinegar in moles per liter (M).
First, let's convert the volumes to liters. The vinegar sample is given as 5.00 mL, so we divide by 1000 to get it in liters: 5.00 mL ÷ 1000 = 0.005 L. Similarly, the NaOH solution used is given as 44.5 mL, so we divide by 1000: 44.5 mL ÷ 1000 = 0.0445 L.
Now we can set up the equation using the known information:
0.100 M NaOH = [Vinegar] / 0.005 L
Rearrange the equation to solve for [Vinegar]:
[Vinegar] = 0.100 M NaOH × 0.005 L
[Vinegar] = 0.0005 mol NaOH
So, the concentration of vinegar in moles per liter is 0.0005 M.
By following these steps, you should be able to solve similar problems on your study guide. Let me know if there's anything else I can help you with!