Question

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Calculate how many milliliters of 0.622 M KOH should be added to 4.90 g of MOPS (see the Structures and pKa Values for Common Buffers table) to give a pH of 7.77. 4.0 23.3 XmL

Answer 1

Firstly, listing some vital values for MOPS from buffer's table:

IUPAC name: 3-morpholinopropane-1-sulphonic acid

Formula: C₇H₁₅NO₄S

Molar mass: 209.2633 g/mol

pKa = 7.20

From the structure, it's clear that it is a mono-basic acid

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Simply, this mono-basic acid can be represented by HA for the sake of this question.

Now, every one mole of KOH added will react with one mole of H⁺ ions derived from HA and reduce [ HA] or [acid] by 1 mole and increase the concentration of its conjugate base (A^-) by 1 mole. See the following chemical equation:

HA_(aq) + KOH_(aq)$\rightleftharpoons$ A^-_(aq) + H₂O_(l) + K⁺_(aq)

Now calculating the number of moles of MOPS from 4.90 g

Number of moles of MOPS = (4.90 g)$\times$(1mol)/(209.2633g) = 0.0234 mol = 23.4 milli-moles

< Conversion of moles to milli-moles for convenience: 0.0234 mol = 0.0234 mol $\times$ (1000 m moles)/1mol) = 23.4 milli moles>

Let the number of milli-moles of KOH added be 'x'.

Then, applying Henderson-Hasselbach's equation:

pH = pK_a + log ([base+x]/[acid-x])

Filling in the values given as well as the number of milli-moles calculated above:

7.77 = 7.20 + log{(23.4+x)/23.4-x)}

=> log{(23.4+x)/23.4-x)} = 0.57

=> (23.4+x)/(23.4-x) = 10^0.57

=> (23.4+x)/(23.4-x) = 3.715

=> 23.4 + x = 86.931 - 3.715x

=> 4.715x = 86.931 - 23.4

=> 4.715x = 63.531

=> x = 13.474

$\therefore$ 13.474 milli-moles of KOH must be added to get the pH of 7.77

Now, finding the volume in mL of KOH solution of 0.622mol/L that contains 13.474 milli-moles

Volume of KOH solution to be added = (13.474 $\times$ 10^-3 moles) $\times$ (1000mL/0.622 moles) = 21.66 mL (Answer)