150 mL of 2.50 M NaOH is added to a 200 mL solution of 1.80 M HF.
1. Write the net ionic equation
2. Determine the pH of the final solution
1. Net ionic equation:
NaOH + HF -> NaF + H2O
In this reaction, NaOH (Sodium hydroxide) reacts with HF (Hydrofluoric acid) to produce NaF (Sodium fluoride) and water.
2. Determining the pH of the final solution is a bit tricky. Since NaF is a salt of a weak acid (HF), it will undergo hydrolysis in water, which will affect the pH. However, without the exact concentration and volume information of the final solution, it is difficult to accurately calculate the pH. So, I'm afraid I don't have the necessary information to provide a specific answer about the pH. But let's just say it will definitely make a splash!
1. To write the net ionic equation, we first need to write out the balanced chemical equation between NaOH and HF:
NaOH + HF -> NaF + H2O
The net ionic equation only includes the species that undergo a change in the reaction. In this case, the Na+ and OH- ions do not change, so they can be omitted. Therefore, the net ionic equation is:
H+ + F- -> HF
2. To determine the pH of the final solution, we need to consider the reaction between NaOH and HF. NaOH is a strong base and HF is a weak acid, so the reaction will result in the complete dissociation of NaOH and only partial dissociation of HF.
First, we can calculate the moles of NaOH and HF:
moles of NaOH = (volume of NaOH solution in L) * (concentration of NaOH in M)
= 0.150 L * 2.50 M
= 0.375 mol
moles of HF = (volume of HF solution in L) * (concentration of HF in M)
= 0.200 L * 1.80 M
= 0.360 mol
Since 1 mole of NaOH reacts with 1 mole of HF, we can see that HF is the limiting reactant in this reaction.
To find the concentration of HF in the final solution, we need to determine the total volume of the solution after reaction. The volume of the final solution is the sum of the volumes of NaOH and HF solutions, which gives:
volume of final solution = volume of NaOH + volume of HF
= 0.150 L + 0.200 L
= 0.350 L
Now, we can calculate the concentration of HF in the final solution:
concentration of HF in the final solution = moles of HF / volume of final solution
= 0.360 mol / 0.350 L
≈ 1.03 M
Since HF is a weak acid, we need to set up an equilibrium expression for the dissociation of HF:
HF ⇌ H+ + F-
The equilibrium constant expression is:
Ka = [H+][F-] / [HF]
Since we have assumed that HF is the only source of H+ ions, [H+] = [HF] and [F-] = [HF] (as HF dissociates into equal amounts of H+ and F- ions). Thus, we can simplify the equation to:
Ka = [H+]^2 / [HF]
Given that the Ka value for HF is approximately 7.2 x 10^-4, we can substitute the concentration of HF into the equation to find the concentration of H+ ions:
7.2 x 10^-4 = [H+]^2 / (1.03 M)
Taking the square root and solving for [H+], we get:
[H+] ≈ 0.026 M
Finally, we can find the pH of the solution by taking the negative logarithm of the [H+] concentration:
pH = -log10(0.026)
≈ 1.59
To write the net ionic equation, we need to first write the balanced molecular equation and then separate the reactants and products into their respective ions.
1. Write the balanced molecular equation:
NaOH + HF → NaF + H2O
2. Separate the reactants and products into ions:
NaOH(aq) + HF(aq) → Na+(aq) + F-(aq) + H2O(l)
Since Na+ and F- do not participate in the reaction, they are spectator ions and can be omitted from the net ionic equation. The net ionic equation is:
OH-(aq) + H+(aq) → H2O(l)
Now, let's move on to determining the pH of the final solution.
To find the pH of the final solution, we need to identify the strong acid and strong base in the reaction. In this case, NaOH is the strong base and HF is the weak acid.
Since NaOH is a strong base, it completely dissociates into Na+ and OH- ions. This reaction results in an excess of OH- ions in the solution.
The OH- ions react with the H+ ions from the weak acid HF to form water. Since the reaction consumes H+ ions, the concentration of H+ ions decreases, leading to a higher pH.
To determine the pH, we need to calculate the concentration of OH- ions and convert it to pOH. Then, we can use the relationship between pH, pOH, and 14 to find the pH.
First, calculate the moles of NaOH added:
Moles of NaOH = volume of NaOH solution (L) * concentration of NaOH (mol/L)
= 0.150 L * 2.50 mol/L
= 0.375 mol
Since NaOH is a strong base, it fully dissociates, meaning there are 0.375 mol of OH- ions in the final solution.
Next, calculate the total volume of the final solution:
Total volume = volume of NaOH solution + volume of HF solution
= 0.150 L + 0.200 L
= 0.350 L
Now, calculate the concentration of OH- ions in the final solution:
OH- concentration = moles of OH- ions / total volume
= 0.375 mol / 0.350 L
= 1.07 mol/L
To find pOH, we take the negative logarithm of the OH- concentration:
pOH = -log10(1.07)
= 0.97
Finally, we can find the pH using the relationship between pH, pOH, and 14:
pH = 14 - pOH
= 14 - 0.97
= 13.03
Therefore, the pH of the final solution is approximately 13.03.
I did the whole thing, punched the wrong button to post the answer and it disappeared. I've thought twice before doing this LOOOOONg problem.
millimoles HF = mL x M = 200 mL x 1.80 M = 360 mmols HF
millimoles NaOH = 150 mL x 2.50 M = 375 mmols NaOH
...................HF + NaOH ==> NaF + H2O
I..................360........0..............0...........0
add.........................375..........................
C..............-360.......-360.........+360...................
E..................0.............15.........+360......................
(NaF) formed = 360 mmols/350 mL = 1.03 M.
(NaOH) remaining = M = millimols/mL = 15 mmols/350 mL = 0.0428 M. You can assume that the pH is determned by the excess NaOH get the pOH from pOH = -log(OH^-) and determine the pH from
pH + pOH = pKw = 14. You know pKw and pOH, solve for pH.
However, I don't know if this is a beginning class or not but the F^- from the NaF will hydrolyze and produce OH^-. It's easy enough to correct for that that in this way. The hydrolysis equation looks like this.
.....................F^- + HOH ==>HF + OH^-
I....................1.03...................0.......0.043
C....................-x.....................+x..........+x
E....................1.03-x.................x.........0.043+x
Kb for F^- = (Kw/Ka for HF) = (x)(0.043+x)/(1.03-x) and solve for x, evaluate x + 0.043 and convert that to pOH, then to pH as above.
Post your work if you get stuck. The question asks for the equation as net ionic and I wrote a molecular equation.Here's the net ionic equation.
HF + OH^- ==> F^- + H2O
You might want to compare the two OH values you get (and pOH/pH) from the remaining NaOH just by itself with the OH^- from the hydrolysis of the NaF.