# K_b for "acetoaminophen" is 4.1xx10^-4. What is [HO^-] for a solution of "acetoaminophen" at 0.910*mol*L^-1 concentration?

Mar 9, 2017

#### Answer:

$\left[H {O}^{-}\right] = 0.019 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

${C}_{8} {H}_{9} {N}_{3} {O}_{2} N H + {H}_{2} O r i g h t \le f t h a r p \infty n s {C}_{8} {H}_{9} {N}_{3} {O}_{2} N {H}^{+} + H {O}^{-}$

We write the expression for base association in the usual way:

${K}_{b} = \frac{\left[{C}_{8} {H}_{9} {N}_{3} {O}_{2} N {H}^{+}\right] \left[H {O}^{-}\right]}{\left[{C}_{8} {H}_{9} {N}_{3} {O}_{2} N H\right]} = 4.1 \times {10}^{-} 4$

If, $\left[H {O}^{-}\right] = x$, then clearly (because of the size of ${K}_{b}$), $\left[{C}_{8} {H}_{9} {N}_{3} {O}_{2} N H\right] \text{ >> } \left[H {O}^{-}\right]$, i.e. $\left[{C}_{8} {H}_{9} {N}_{3} {O}_{2} N H\right] \text{ >> } x$. Given that $\left[H {O}^{-}\right] = x$, $x = \left[{C}_{8} {H}_{9} {N}_{3} {O}_{2} N {H}^{+}\right]$, and $\left[{C}_{8} {H}_{9} {N}_{3} {O}_{2} N\right] = \left(0.910 - x\right)$

i.e., so if ${K}_{b} = \frac{{x}^{2}}{0.910 - x}$

${K}_{b} \cong \frac{{x}^{2}}{0.910}$, this is an approximation, which we must justify later.

${x}_{1} = \sqrt{{K}_{b} \times 0.910} = \sqrt{4.1 \times {10}^{-} 4 \times 0.910} = 0.019$

Since we got an approximation for $x$, we can recycle this value thru the equilibrium expression, and see if our estimate of ${x}_{2}$ varies markedly:

x_2=sqrt(K_bxx (0.910-0.019))=sqrt(4.1xx10^-4xx(0.910-0.019))=0.019*mol*L^-1

Our approximation was indeed reasonable. And thus $x = \left[H {O}^{-}\right] = 0.019 \cdot m o l \cdot {L}^{-} 1$. What is $p H$ for this solution?