1. 50.0 mL of 0.00040 M Mg(NO3)2 are mixed with 150.0 mL of 0.00040 M NaOH....

Question

Question

1. 50.0 mL of 0.00040 M Mg(NO3)2 are mixed with 150.0 mL of 0.00040 M NaOH. Determine whether a precipitate will form.

2. 150.0 mL of 0.00040 M Mg(NO3)2 are mixed with 50.0 mL of 0.00040 M NaOH. Determine whether a precipitate will form.

science chemistry

Answer 1

Answer #1

mg (NO3)2 + 2NaOH → Mg (OH)2 + 2NaNO3] Concentration of Mg2+ =- et (500 ml) * (400x2104 M) Total voleimne = (5000 + 150.0) mb

Question 2 is same as 1 just they have reversed the value of volume.

The Qsp will come out 3×10^ -12 .

Since this value is less than Ksp precipitate will not occur.

Answer 2

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