Question

(LARGE MULTIPLE PART QUESTION - Help with any of the parts is greatly appreciated!) The antibiotic penicillin has a thiazolidine ring in its structure. The compound thiazolidine, C3H7NS, is a molecular weak base with a pKb of 7.69?

A. What is the formula of the related conjugate acid, thiazolidinium?

B. What is the pH and %dissociation of a 0.0876Msolution of thiazolidine?

C. What is the pH and %dissociation of a 0.0543Msolution of thiazolidinium chloride, HC3H7NSCl?

D. What is the pH of a buffer solution containing both 0.0876 MC3H7NS and 0.0543M HC3H7NSCl?

Answer 1

Part A. The formula of the conjugate base is `"C"_3"H"_8"NS"^"+"`.
Part B. pH = 9.63; `α = 0.0482 %`

Explanation:

Your question is so long that I will break the answer into separate parts.

A. The formula of the conjugate base

The equation for the equilibrium is

`underbrace("C"_3"H"_7"NS")_color(red)("base") + underbrace("H"_2"O")_color(red)("acid") ⇌ underbrace("C"_3"H"_8"NS"^"+")_color(red)("conj. acid") + underbrace("OH"^"-")_color(red)("conj. base")`

The thiazolidinium ion has one more proton than thiazolidine. Its formula is `"C"_3"H"_8"NS"^"+"`.

B. pH and % dissociation of thiazolidine solution

For convenience, let's write the formula of the base as `"B"`.

Then we can write an ICE table as

`color(white)(mmmmmmmll)"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"`
`"I/mol·L"^"-1": color(white)(ml)0.0876 color(white)(mmmmm)0color(white)(mmml)0`
`"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmll)"+"xcolor(white)(mml)"+"x`
`"E/mol·L"^"-1": color(white)(m)"0.0876-"xcolor(white)(mmmml)xcolor(white)(mmm)x`

`"p"K_text(b) = 7.69`

`K_text(b) = 10^"-7.69" = 2.04 × 10^"-8"`

`K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = x^2/("0.0876-"x) = 2.04 × 10^"-8"`

Check for negligibility

`0.0876/(2.04 × 10^"-8") = 4.3 × 10^6 > 400`. ∴ `x ≪ 0.0876`

The equilibrium constant expression becomes

`x^2/0.0876 = 2.04 × 10^"-8"`

`x^2 = 0.0876 × 2.04 × 10^"-8" = 1.789 × 10^"-9"`

`x = 4.229 ×10^"-5"`

`["OH"^"-"] = xcolor(white)(l) "mol/L" = 4.229 ×10^"-5"color(white)(l)"mol/L"`

`"pOH" = "-log"["OH"^"-"] = "-log"(4.229 ×10^"-5") = 4.37`

`"pH = 14.00 - pOH = 14.00 - 4.37 = 9.63"`

`α = (["OH"^"-"])/(["B"]_0) × 100 % = (4.229 ×10^"-5")/0.0876 × 100 % = 0.0482 %`

Answer 2

C. pH = 3.79; `α = 0.303 %`; D. pH = 6.52

Explanation:

C. pH and % dissociation of thiazolidinium ion

Set up an ICE table.

`color(white)(mmmmmmml)"BH"^"+" + "H"_2"O" ⇌ "B" + "H"_3"O"^"+"`
`"I/mol·L"^"-1": color(white)(ml)0.0534 color(white)(mmmmmll)0color(white)(mmml)0`
`"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x`
`"E/mol·L"^"-1": color(white)(m)"0.0534-"xcolor(white)(mmmmm)xcolor(white)(mmm)x`

`K_text(a) = K_text(w)/K_text(b) = (1.00 ×10^"-14")/(2.04 × 10^"-8") = 4.90 × 10^"-7"`

`K_text(a) = (["B"]["H"_3"O"^"+"])/(["BH"^"+"]) = x^2/("0.0534-"x) = 4.90 × 10^"-7"`

Check for negligibility

`0.0534/(4.70 × 10^"-7") = 1.1 × 10^5 > 400`. ∴ `x ≪ 0.0534`

The equilibrium constant expression becomes

`x^2/0.0534 = 4.90 × 10^"-7"`

`x^2 = 0.0534 × 4.90 × 10^"-7" = 2.618 × 10^"-8"`

`x = 1.618 ×10^"-4"`

`["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 1.618 ×10^"-4"color(white)(l)"mol/L"`

`"pH" = "-log"["H"_3"O"^"+"] = "-log"(1.618 ×10^"-4") = 3.79`

`α = (["H"_3"O"^"+"])/(["B"]_0) × 100 % = (1.618 ×10^"-4")/0.0534 × 100 % = 0.303 %`

Part D. pH of buffer

The equation for the equilibrium is

`color(white)(mmmmmmmll)"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"`
`"E/mol·L"^"-1": color(white)(m)0.0876 color(white)(mmmll)0.0543`

The Henderson-Hasselbalch equation is

`"pOH = p"K_text(b) + log((["BH"^"+"])/(["B"]))`

∴ `"pOH" = 7.69 + log(0.0543/0.0876) = 7.69 +log0.6199 = 7.69 - 0.208 = 7.48`

`"pH = 14.00 - pOH = 14.00 - 7.48 = 6.52"`