Given stock solutions of 25 mM

Given stock solutions of 25 mM nitrophenyl phosphate, and 1.5 Mtris,pH 7.2, 7.5 mM MgCl2. If you mix 0.1ml of the nitrophenylphosphate, 0.7 ml of the tris solution and 0.2 ml of water to bringthe total volume to 1 ml, what is the final concentrations of eachof the components in the final solution? Show work

Nitrophenyl phosphate in mM

Tris in M

Magnesium chloride in mM


In the given question, we are adding 0.1ml of the nitrophenylphosphate, 0.7 ml of the tris solution and 0.2 ml of water to bringthe total volume to 1 ml.

Total volume = 1 ml

We need to first calculate the no of mmols in 0.1ml ofthe nitrophenyl phosphate

Molarities of Stock solutions of nitrophenyl phosphate, tris,pH7.2 and MgCl2 will give us the No. of moles of eachspecies in 1 Lt of Volume.

Molarity= No. of moles dissolved in 1 Lt of solution

So, 25 mM nitrophenyl phosphate will contain 25 mmols ofnitrophenyl phosphate in 1 Lt of volume.

1 lt(1000 ml) of 25 mM nitrophenyl phosphate solution will contain = 25mmols of nitrophenyl phosphate

0.1ml of 25mM nitrophenyl phosphate solution will contain= 0.1 * 25 /1000 =25* 10-4 mmoles ofnitrophenyl phosphate


1.5 Mtris(pH 7.2) will contain 1.5 mols (1500 mmols) of tris in1 Lt of volume.

0.7 ml of 1.5 M tris(pH 7.2) solutionwill contain= 0.7* 1.5 /1000 = 1.05* 10-3 molesoftris

So we are adding 25* 10-4 moles ofnitrophenylphosphate and 1.05* 10-3 molsoftris and making the total volume 1 ml by diluting withwater.

Now we will calculate the molarity,

For nitrophenylphosphate :

1 ml of total volume contains = 25* 10-4 mmolesofnitrophenyl phosphate

1000ml of total volume will contain = 25* 10-1mmolesofnitrophenyl phosphate or 2.5 mmols

Concentration of nitrophenyl phosphate = 2.5 mmoles/Lt= 2.5mM


1 mlof total volume contains = 1.05* 10-3 moles oftris

1000ml of total volume will contain = 1.05 moles oftris

Concentration of tris = 1.05 moles/Lt = 1.05M

Since we havenot added MgCl2 in given question, itsConcentration will be zero.

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