I want to make 1.0 liter of a buffer that will keep the pH of a solution at 9.50

Question

I want to make 1.0 liter of a buffer that will keep the pH of a solution at 9.50. If I start by dissolving 15 g of piperidine in water, how many milliters of either 0.5 M HCl, or 0.5M NaOH will I have to add to this solution to make my buffer? Piperidine, pK_b =2.89, Molecular formula: C5H11N Calculate the theoretical potential of the cell Zn/ZnSO_4(0.5M), PbSO_4(saturated)/Pb using concentration; using activity. PbSO_4(s) + 2e^- tensor Pb(s) + SO_4^2- E^0 = -0.350V Zn^2+ + 2e^- tensor Zn(s) E^0 = -0.763V A 2.16 g sample containing both Fe and V was dissolved under certain conditions and diluted to 500.00 mL. A first 50.00 mL aliquot was taken and passed through a Walden reductor to form Fe^2+ and VO^2+ ions. The titration of this solution required 16.84 mL of 0.1000 M Ce^4+ to reach end point. A second 50.00 mL aliquot was passed through a Jones reductor to form Fe^2+ and V^2+ ions. The titration of the second solution required 44.26 mL of 0.1000 M Ce^4+ solution to reach an end point. Calculate the percentage of Fe_2 0_3 and V_2 O_5 in the sample. In the first titration, In the second titration, Derive a titration curve for 20.00 mL of 0.1000 M Ca^2+ with 0.500 M EDTA in a solution buffered to pH 10.0. Determine the pCa values after the addition of 20.00, 40.00 and 60.00 mL of EDTA. K_MY(CaEDTA) = 5.00 * 10^10.

Explanation / Answer

Hi, as per Chegg question policy, I will be answering question 1 for you.

A buffer will be composed of piperidine, which is a base, and its conjugate acid. For that, we will have to add HCl to it to form its conjugate acid.

According to the Henderson Hasselbach equation,

pOH = pKb + log ([conjugate acid]/[base])

Here pOH = 14-pH = 4.5

pKb = 2.89

Now, number of moles of piperidine added= weight of piperidine added / molar mass of piperidine = 15/ 85.15 = 0.176 mol

Now, let's assume we add x moles of HCl to it. It will react stoichiometrically with piperidine to form x moles of the conjugate acid. (0.176-x) moles of piperidine base will be left in solution.

So, [conjugate acid] = x

[base]= 0.176-x

Substituting these values in the HH equation we get

4.5 = 2.89 + log (x/ (0.176-x))

log (x/ (0.176-x)) = 1.61

So, 10log (x/ (0.176-x)) = 101.61 = 40.73

x/ (0.176-x) = 40.73 (Since 10logy = y)

Which gives x = 0.1718 mol

Now, molarity of HCl solution = 0.5M

So, for mLs of HCl required,

0.5 = 0.1718/V

V = 0.343 L = 343mL

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