It is considered a DUI if the driver’s Blood Alcohol Content level is at or above .08.
1 answer:
You might be interested in
Explanation:
Step 1:
A good first step for a problem like this is to write down the chemical formula and balance it.
It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:
CH3COOH + NaOH ------> NaCH3COO + H2O
All of the constituents come out to a value of 1, conveniently.
Step 2:
Since all of our stoichiometric coefficients are one, we can use a shortcut to answer this equation. I don't know if it has a name, but I just call it the titration formula. It goes something like this:
M1 * V1 = M2 * V2
M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.
So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.
Our V1 is going to be the initial volume of the sample, which was 10.5 mL
Our V2 is going to be 19.13, which is the volume when we're finished.
It's clear that we don't know M2, so let's find it.
Keep in mind that it's easier to convert to liters pretty much always, so I've done that by dividing the mL values each by 1000.
Using some algebra, we can see that we now have:
0.460 M * 0.0105 L = x M * 0.01913 L
Which goes to:
The pH of the solution is basically the negative logarithm of the concentration of hydrogen ions or H+. In equation form, pH = -log[H+]. It could also be in terms of the concentration of hydroxide ions or OH- as pOH, where pOH = -log[OH-]. The sum of pH and pOH is 14. These are the important equations to know when it comes to equilibrium pH problems.
KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be
CN- + H2O ⇔ HCN + OH-
The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.
CN- + H2O ⇔ HCN + OH-
I 0.2 m ∞ 0 0
C -x ∞ +x +x
-----------------------------------------------------------
E 0.2-x +x +x
The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:
![K_{H} = \frac{ K_{W} }{ K_{A} } = \frac{[HCN][OH-]}{[CN-]}](https://tex.z-dn.net/?f=%20K_%7BH%7D%20%3D%20%5Cfrac%7B%20K_%7BW%7D%20%7D%7B%20K_%7BA%7D%20%7D%20%3D%20%5Cfrac%7B%5BHCN%5D%5BOH-%5D%7D%7B%5BCN-%5D%7D%20)
where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,
![K_{H} = \frac{ 1x10^-14}{ 6.2x10^-10 } = \frac{[X][X]}{[0.2-X]}](https://tex.z-dn.net/?f=K_%7BH%7D%20%3D%20%5Cfrac%7B%201x10%5E-14%7D%7B%206.2x10%5E-10%20%7D%20%3D%20%5Cfrac%7B%5BX%5D%5BX%5D%7D%7B%5B0.2-X%5D%7D)
x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,
pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25
KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be
CN- + H2O ⇔ HCN + OH-
The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.
CN- + H2O ⇔ HCN + OH-
I 0.2 m ∞ 0 0
C -x ∞ +x +x
-----------------------------------------------------------
E 0.2-x +x +x
The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:
where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,
x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,
pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25