pH calculation - adding distilled water after titration
So I presume you can work out the initial H-H equation. I got pH 5.8? might be worth checking.
Then use the equation $$c_1 \cdot V_1 = c_2 \cdot V_2$$
Where $c_1 = [\ce{H+}]$

$[\ce{H+}] = 10^{-\mathrm{pH}}$

so I got $[\ce{H+}] =10^{-5.8} = 1.58489 \times 10^{-6}$

Now we can rearrange our original equation: $(c_1 \cdot V_1)/V_2 = c_2$

$c_1=1.585\times10^{-6}$

$V_1=120\ \mathrm{ml}=0.12\ \mathrm L$

$V_2=(120\ \mathrm{ml}+880\ \mathrm{ml})= 1\ \mathrm L$

$c_2=1.902\times10^{-7}$

$\mathrm{pH}=-\log[\ce{H+}]$

$\mathrm{pH}= -\log(1.902\times10^{-7})$

$\mathrm{pH}= 6.72$

Hope this helps, Its been a long time since I did anything like this so let me know how it goes and if this is correct. cheers.