It will be around 7. 13.3 is very alkaline, on par with a strong alkaline such as NaOH.
First, you need your Kw value. Just looked in my notes, and for 60 degrees it is 5.60 x 10-14 mol2 dm-6. Check the question and use the value from there. I'll use the 60 degrees value for this example.
Kw = [H+][OH-]
so [H+][OH-]
[H+](squared) = 5.60 x 10-14
[H+] = square root of 5.60 x 10-14
[H+] = 2.366 x 10-7 mol dm-3
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pH = -log10 [H+]
so log 2.366 x 10-7 = 6.62
pH = 6.62
Thats for 60 degrees, I think the Kw value for 50 degrees will be different. Hope that helps!