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calculate the ph and concentration of a buffer made with 90 mL of .4 M PIPES...

calculate the ph and concentration of a buffer made with 90 mL of .4 M PIPES acid and 30 ml of .6M NaOH and then diluted to 400 mL with water

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Answer #1

Solution :-

Lets first calculate the moles of the NaOH and PIPES

Moles = molarity * volume in liter

Moles of of PIPES = 0.4 mol per L * 0.090 L = 0.036 mol

Moles of NaOH = 0.6 mol per L * 0.030 L = 0.018 mol NaOH

So after the reaction 0.018 mol es of the PIPES will be converted to its conjugate base

So moles of PIPES after reaction = 0.036 mol – 0.018 mol = 0.018 mol

Moles of conjugate base = 0.018 mol

Total volume = 90 ml + 30 ml + 400 ml = 520 ml = 0.520 L

New molarities of the PIPES and conjugate base are as follows

Molarity = moles / volume in liter

[PIPES] = 0.018 mol / 0.520 L = 0.0346 M

[conjugate base ] = 0.018 mol / 0.520 L = 0.0346 M

Pka of the PIPES = 6.76

Using the Henderson equation we can calculate the pH of the solution

pH = pka + log [conj. Base ]/[acid]

pH= 6.76 + log [0.0346 / 0.0346]

pH= 6.76

So the pH of the solution is 6.76

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