Al(NO3)3 ==> Al^+3 + 3NO3^-
There are three (3) nitrate ions per mol Al(NO3)3; therefore, 0.9 M Al(NO3)3 x [3 NO3^-/1 mol Al(NO3)3] = 0.9 M x 3/1 = ? M NO3^- ion
In the question Consider a 0.90 M Al(NO3)3 solution. This solution has a nitrate ion concentration of . . . .
How do you figure the nitrate ion concentration?
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have absolutly no idea
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