Alevel chemistry kw

was the final answer 12.57???

I worked out the moles of both and then did the ice box. x2 the moles of h2s04 because its a 1;2 ratio and then got excess moles of koh which was 1.8 something. convert that to concentration and then use kw which i just assumed was 1x10-14.

i) 0.48 x 0.03 = 0.0144 mol

ii) 0.018 x 0.35 = 6.3 x10-3 mol

iii) 0.0144 - 2(6.3x10-3) = 1.8x10-3-----> 2 x 6.3x10-3 because sulfuric acid is diprotic

iv) 1.8x10-3/ 0.048 = 0.0375 moldm-3

v) 1x10-14/ 0.0375 = 2.667x10-13 moldm-3 = [H ] ---> Then -log[H ] = 12.57

i think youre doing an old paper? correct me if im wrong, but they will most likely tell you.

For my exam board (AQA) you are supposed to know what kw is at 298k. If they do not give you a temperature, I think you assume that it was done under standard conditions (which is 298k). Unless you are supposed to know multiple kw values, they would have to tell you the different temperature and give you a new kw value for that temperature.

if you subbed the moles, it might be wrong because you didnt convert to concentration

for the final part how did you know kw was = to 10^-14

i) 0.48 x 0.03 = 0.0144 mol
ii) 0.018 x 0.35 = 6.3 x10-3 mol
iii) 0.0144 - 2(6.3x10-3) = 1.8x10-3-----> 2 x 6.3x10-3 because sulfuric acid is diprotic
iv) 1.8x10-3/ 0.048 = 0.0375 moldm-3
v) 1x10-14/ 0.0375 = 2.667x10-13 moldm-3 = [H ] ---> Then -log[H ] = 12.57