How do I prepare a 10ml mixture of 0.0200M Cr(NO3)3 and 0.0752M Co(NO3)2 if I have...

How do I prepare a 10ml mixture of 0.0200M Cr(NO₃)₃ and 0.0752M Co(NO₃)₂ if I have a stock solution of 25ml of 0.0500M Cr(NO₃)₃ and another stock solution of 25ml of 0.1880M Co(NO₃)₂?

Expert Solution

Total volume of the mixture to be prepared, Vt = 10 mL

Given the concentration of Cr(NO3)3 in the mixture = 0.0200 M

Hence millimoles of Cr(NO3)3 in the mixture = Volume x concentration = 10 mL x 0.0200M = 0.200 millimol

Given the concentration of Co(NO3)2 in the mixture = 0.0752M

Hence millimoles of Co(NO3)2 in the mixture = Volume x concentration = 10 mL x 0.0752M = 0.752 millimol

Let the volume of Cr(NO3)3 taken from 0.0500M Cr(NO3)3 stock solution be 'V' mL

Hence millimoles of Cr(NO3)3 taken from stock solution = millimoles of Cr(NO3)3 in the mixture

=> V mL x 0.0500M = 0.200 millimol

=> V = 0.200 millimol / 0.0500M = 4 mL

Hence volume of 0.0500M Cr(NO3)3 required from stock solution = 4 mL

Hence volume of 0.1880M Co(NO3)2 required = Vt - 4 mL = 10 mL - 4 mL = 6 mL

Hence we need to add 4 mL of 0.0500M Cr(NO3)3 stock solution and 6 mL of 0.1880M Co(NO3)2 stock solution to prepare the mixture. (answer)

queen_honey_blossom answered 2 years ago

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