Question

How would you make a 50.0 mM MOPS buffer (pKa=7.15) at pH=7.00 using MOPS acid (324.4 g/mol) and MOPS base (346.3 g/mol)? Provide a recipe, indicating the following:
Grams of MOPS base:
Grams of MOPS acid:
Total volume:

Answer 1

Question#6042

How would you make a 50.0 mM MOPS buffer (pKa=7.15) at pH=7.00 using MOPS acid (324.4 g/mol) and MOPS base (346.3 g/mol)? Provide a recipe, indicating the following:

Grams of MOPS base:

Grams of MOPS acid:

Total volume:

Solution:

$$\mathrm{ph} = \mathrm{p} K_{\mathrm{A}} - \lg \frac{[\mathrm{acid}]}{[\mathrm{salt}]}$$ $$7 = 7,15 - \lg \frac{[\mathrm{MOPS}_{\mathrm{A}}]}{[\mathrm{MOPS}_{\mathrm{B}}]}$$ $$\lg \frac{[\mathrm{MOPS}_{\mathrm{A}}]}{[\mathrm{MOPS}_{\mathrm{B}}]} = 0,15$$ $$\frac{[\mathrm{MOPS}_{\mathrm{A}}]}{[\mathrm{MOPS}_{\mathrm{B}}]} = 1,413$$ $$[\mathrm{MOPS}_{\mathrm{A}}] = 1,413 \times [\mathrm{MOPS}_{\mathrm{B}}]$$ $$\theta = [\mathrm{MOPS}] \times \mathrm{V}$$

For 1L of $50.0\,\mathrm{mM}$ solution: $\theta (\mathrm{MOPS}) = [\mathrm{MOPS}]$, because $(\mathrm{V} = 1)$

$$\theta = \frac{\mathrm{m}}{\mathrm{M}_{\mathrm{r}}}$$ $$\theta (\mathrm{MOPS}_{\mathrm{A}}) + \theta (\mathrm{MOPS}_{\mathrm{B}}) = 0,05$$ $$\theta (\mathrm{MOPS}_{\mathrm{A}}) = 1,413 \times \theta (\mathrm{MOPS}_{\mathrm{B}})$$ $$1,413 \times \theta (\mathrm{MOPS}_{\mathrm{B}}) + \theta (\mathrm{MOPS}_{\mathrm{B}}) = 0,05$$ $$\theta (\mathrm{MOPS}_{\mathrm{B}}) = \frac{0,05}{2,413} \approx 0,021\,\mathrm{mol}$$ $$\mathrm{m} (\mathrm{MOPS}_{\mathrm{B}}) = \theta \times \mathrm{M}_{\mathrm{r}} = 0,021 \times 346,3 = 7,27\,\mathrm{g}$$ $$\theta (\mathrm{MOPS}_{\mathrm{A}}) = 1,413 \times \theta (\mathrm{MOPS}_{\mathrm{B}}) = 1,413 \times 0,021 \approx 0,030\,\mathrm{mol}$$ $$\mathrm{m} (\mathrm{MOPS}_{\mathrm{A}}) = \theta \times \mathrm{M}_{\mathrm{r}} = 0,030 \times 324,4 = 9,73\,\mathrm{g}$$

Answer:

Grams of MOPS base: 7.27

Grams of MOPS acid: 9.73

Total volume: 1L