Need help!
Calculate the molar concentration of OH- in a 0.724M solution of hypobromite ion (BrO-; kB= 4.0 x 10^-6) what is the pH of this solution?
I got as far as kB = [OH-][HBrO]/[BrO-] please help me with the rest.
You did te hard part. You just quit too soon. Here is the equation.
................BrO^- + HOH ==> HBrO + OH^-
I..............0.724.......................0.............0
C..............-x...........................x..............x
E...........0.724-x......................x..............x
You have written the Kb expression. Just substitute what each is and solve for x = (OH^-) and convert to pH
Okay so I got. 4.0 x10^-6 = x^2/0.724m-x
Right??
Right but not m. If you want to put the unit in it would be M. Note that m stands for molality and M stands for molarity.
Ah, chemistry! The science that makes us question our sanity. Now, let's see if we can decipher this equation and find a solution!
You've already made a good start with the equation for the equilibrium constant. Here's what we need to do next:
1. Write out the chemical equation for the dissociation of hypobromite ion (BrO-) in water.
2. Determine the initial concentration of hypobromite ion.
3. Express [OH-] and [HBrO] in terms of x (the change in concentration).
4. Substitute these values into the equation for the equilibrium constant.
5. Solve for x.
6. Determine the molar concentration of OH- and pH!
Okay, let's dive into it, brave chemist!
Step 1: The equation for the dissociation of hypobromite ion (BrO-) in water is as follows:
BrO- + H2O ⇌ OH- + HBrO
Step 2: The initial concentration of hypobromite ion is given as 0.724M. So we have [BrO-] = 0.724M.
Step 3: We express [OH-] and [HBrO] in terms of x. Let's assume that [OH-] and [HBrO] at equilibrium are equal to x M.
Step 4: Now, let's substitute these values into the equation for the equilibrium constant:
kB = [OH-][HBrO]/[BrO-]
kB = (x)(x) / 0.724
Step 5: Solve for x.
x^2 = kB * 0.724
Take the square root of both sides to find x.
x = √(kB * 0.724)
Step 6: Phew, we're almost there! Use the molar concentration of OH- to determine the pH using the formula: pH = -log10([H+]), since [H+] and [OH-] are inversely related.
[H+] = [OH-]
pOH = -log10(x)
pH = 14 - pOH
Now, my little scientist, plug in the value of x you calculated and find that elusive pH!
Remember, chemistry jokes might not improve your pH, but they'll definitely improve your mood!
To calculate the molar concentration of OH- in a solution of hypobromite ion (BrO-), you already have the expression kB = [OH-][HBrO]/[BrO-]. We can rearrange this equation to solve for [OH-]:
kB = [OH-][HBrO] / [BrO-]
[OH-] = (kB * [BrO-]) / [HBrO]
Now, let's substitute the known values into the equation:
kB = 4.0 x 10^-6
[BrO-] = 0.724 M
To find the concentration of HBrO, we can use the fact that the solution is 0.724 M. In aqueous solutions, hypobromite (BrO-) acts as a weak base and can accept a proton (H+) to form hypobromous acid (HBrO). So, we can assume that the concentration of HBrO is also 0.724 M.
Substituting these values into the equation:
[OH-] = (4.0 x 10^-6 * 0.724) / 0.724
[OH-] = 4.0 x 10^-6 M
Therefore, the molar concentration of OH- in the solution is 4.0 x 10^-6 M.
To determine the pH of the solution, we can use the fact that pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration (H+).
pOH = -log10([OH-])
pOH = -log10(4.0 x 10^-6)
pOH = -[-6log10(4.0)]
pOH = -[-6 + log10(4.0)]
pOH = -[-6 + 0.6021]
pOH ≈ 6.6021
Since pH + pOH = 14 (at 25 °C), we can subtract the pOH from 14 to find the pH:
pH = 14 - pOH
pH = 14 - 6.6021
pH ≈ 7.3979
Therefore, the pH of this solution is approximately 7.40.