\(n_{H_2SO_4.5H_2O}=\dfrac{12,5}{98+5.18}=0,066\left(mol\right)\)
Khi hòa tan \(H_2SO_4.5H_2O\) vào nước:
\(H_2SO_4.5H_2O\rightarrow H_2SO_4+5H_2O\)
0,066--------------> 0,066
Ta coi \(H_2SO_4\) điện li hoàn toàn 2 nấc:
\(H_2SO_4\rightarrow2H^++SO_4^{2-}\)
0,066----->0,132--->0,066
\(\left[H^+\right]=\dfrac{0,132}{0,2}=0,66M\)
\(\left[SO_4^{2-}\right]=\dfrac{0,066}{0,2}=0,33M\)