Let's start by finding the number of moles of NaOHNaOH put into the solution, using the concentration formula:
c=(n)/vc=nv
cc=conc in mol dm^-3moldm−3
nn=number of moles
vv=volume in litres ( dm^3dm3)
50.0 ml=0.05 dm^(3)=v50.0ml=0.05dm3=v
0.2 times 0.05=n0.2×0.05=n
n=0.01 mol n=0.01mol
And to find the number of moles of HFHF:
c=(n)/vc=nv
0.5=(n)/0.020.5=n0.02
n=0.1n=0.1
NaOH (aq) + HF(aq) -> NaF(aq) + H_2O(l)NaOH(aq)+HF(aq)→NaF(aq)+H2O(l)
We form 0.1 mol of NaFNaF in the resulting 70ml solution after the reaction has gone to completion.
Now, NaFNaF will be dissociated in the solution, and the fluoride ion, F^(-)F− will act as a weak base in the solution(We will come back to this).
So now is a good time to set up an ICE table to find the amount of OH^-OH− ions it forms, but we first need to know the concentration of the NaFNaF, as concentrations are used in the ICE table.
c=(n)/vc=nv
c=(0.1/0.07)c=(0.10.07)
c approx 0.143 mol dm^-3c≈0.143moldm−3 of NaFNaF ( =[F^(-)])=[F−])
The reaction of the flouride ion and the consequent concentration changes are:
"color(white)(mmmmm)F^(-)(aq) + H_2O(l) -> HF(aq) + OH^(-)(aq)mmmmmF−(aq)+H2O(l)→HF(aq)+OH−(aq)
"Initial:"color(white)(mm)0.143color(white)(mmm)-color(white)(mmmm)0color(white)(mmmmll)0Initial:mm0.143mmm−mmmm0mmmmll0
"Change:"color(white)(iim)-xcolor(white)(mmm)-color(white)(mmm)+xcolor(white)(mmll)+xChange:iim−xmmm−mmm+xmmll+x
"Eq:"color(white)(mmm)0.143-xcolor(white)(mii)-color(white)(mmmm)xcolor(white)(mmmmll)xEq:mmm0.143−xmii−mmmmxmmmmllx
The K_bKb expression for the fluoride ion would be:
K_b=([OH^(-)] times [HF])/([F^(-)])Kb=[OH−]×[HF][F−]
But how do we know the K_bKb for the fluoride ion, which we touched upon earlier?
Well, as we are given that the reaction occurs at 25 degrees Celcius the following property applies:
(K_b) times (K_a)= 1.0 times 10^-14(Kb)×(Ka)=1.0×10−14
For an acid/base pair- and we happen to have the pair of HFHF and F^(-)F−!
Hence:
K_b=(1.0 times 10^-14)/(7.2 times 10^(-4)Kb=1.0×10−147.2×10−4
K_b (F^(-)) approx 1.39 times 10^(-11)Kb(F−)≈1.39×10−11
So now we can create a K_bKb expression and solve for xx to find the concentration of OH^(-)OH−, and thus find the pOHpOH and then consequently the pHpH.
1.39 times 10^(-11)=(x^2)/(0.143-x)1.39×10−11=x20.143−x
K_bKb is small, so the small x approximation gives:
1.39 times 10^(-11)=(x^2)/(0.143)1.39×10−11=x20.143
x=[OH^(-)] = sqrt(K_bcdot0.143)x=[OH−]=√Kb⋅0.143
=1.4095 xx 10^(-6) approx 1.41 times 10^(-6)=1.4095×10−6≈1.41×10−6
Now:
pOH=-log[OH^(-)]pOH=−log[OH−]
pOH=-log[1.41 times 10^(-6)]pOH=−log[1.41×10−6]
pOH approx 5.85pOH≈5.85
And as we are at 25 degrees this property applies:
pH+pOH=14pH+pOH=14
Hence,
pH=14-5.85pH=14−5.85
color(blue)(pH= 8.15)pH=8.15
