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2Naoh + h2so4-> na2so4 + 2h2o
liczymy liczbe moli naoh i h2s4
0,3*0,2=0,06
0,1*0,1=0,01
substraty reaguja w stosunku 2:1
2mole naoh-1 mol h2so4
x= 0,1
x=0,2 mola naoh
naoh w nadmiarze
zostaje 0,4 mola naoh
v=400 dm3 sumujemy cala objetosc
cm=0,04/0,4=0,1
pOH=1
wiec
Ph=13