Since 1 M solution is more concentrated than 0.25 M solution, we will have to dilute the higher strength solution (with distilled water) to obtain a lower strength solution. The following equation will help us decide on the dilution:
C1V1 = C2V2
where, C1 and C2 are concentrations (in M) of solutions and V1 and V2 are volume of these solutions, respectively.
Here, C1 = 1 M, C2 = 0.25 M, V1 is unknown and...
Since 1 M solution is more concentrated than 0.25 M solution, we will have to dilute the higher strength solution (with distilled water) to obtain a lower strength solution. The following equation will help us decide on the dilution:
C1V1 = C2V2
where, C1 and C2 are concentrations (in M) of solutions and V1 and V2 are volume of these solutions, respectively.
Here, C1 = 1 M, C2 = 0.25 M, V1 is unknown and V2 = 10 ml.
Thus, V1 = (C2V2)/C1 = (0.25 x 10)/1 = 2.5 ml
Thus, 2.5 ml of 1 M HCl solution will be required to prepare a 10 ml solution of 0.25 M HCl. Since the final solution volume is 10 ml, the other 7.5 ml (after adding 2.5 ml 1 M HCl) will be made up of distilled water.
Thus, to prepare 10 ml, 0.25 M HCl solution, add 7.5 ml distilled water to 2.5 ml, 1 M HCl.
Hope this helps.
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