A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c

Question

sertanlavr [38]

4 years ago

14

A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c

oncentration of the nitrate ion?

Chemistry

1 answer:

pshichka [43] · Answer 1 · 2021-11-16T18:53:09+03:00

pshichka [43]4 years ago

6 0

<span>Answer: Moles Ca(NO3)2 = 100 x 0.250 / 1000 = 0.025 Ca(NO3)2 >> Ca2+ + 2NO3- Moles NO3- = 2 x 0.025 = 0.05 Moles HNO3 = 400 x 0.100 / 1000 = 0.04 Total moles = 0.05 + 0.04 = 0.09 Total volume = 500 ml = 0.500 L M = 0.09 / 0.500 = 0.18</span>