Question

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

Answer 1

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.