Use activities to calculate the pH of each of the following solutions, being sure to use $\mathrm{\alpha}$ values and the extended Debye-Huckel equation.

The first one the concentration is $0.020 \mathrm{M}$ $\ce{NaCl}$ with the answer being : $\mathrm{pH}$ = $6.99$

The second one the concentration is $0.10 \mathrm{M}$ $\ce{NaCl}$ with the answer being : $\mathrm{pH}$ = $6.98$

The thing is i have no idea how to find the activity coefficient of $[\ce{H}^+]$

I know ionic strength $u$ is calculated as $\mathrm{\frac{1}{2}\left(\sum C_iZ_i^2\right)}$

but how does $[\ce{H+}]$ work into the charge balance?

I also know that the activity coefficient is calculated as


The best I have is: $$\ce{Na+ + Cl- + H+ + OH-}$$

and that raises the question of how do I know what $[\ce{H+}]$ is and $[\ce{OH-}]$ for calculating the ionic strength? Am I approaching this wrong? Where do I start?

  • $\begingroup$ $\ce{NaOH + HCl -> NaCl + H2O}$. It is possible to find the $\ce{[H+]}$ from the pH by $10^{-pH}$ and $\ce{[OH-]}$ by $10^{-pOH}$ $\endgroup$ – Jun-Goo Kwak Apr 1 '14 at 12:13
  • $\begingroup$ I'm sorry maybe i worded the question wrong. We're solving for the pH and i already know the answer. I just don't know how to arrive at it. $\endgroup$ – John Snow Apr 1 '14 at 17:15
  • $\begingroup$ Is it $u$ ($u$) or is it $\mu$ ($\mu$)? $\endgroup$ – Jan Nov 22 '16 at 22:56

Let's first look at ionic strength:

Ionic strength is a measure of the total ion concentration in solution, however, species with higher charge can also have elctrostatic interactions with other species.

$$\mu = \frac{1}{2} \sum_i C_iZ_i^2 $$

What is the strength of 0.02 M NaCl solution?

$$\mu = \frac{1}{2}([\ce{Na+}]z^2 + [\ce{Cl-}]z^2) = \frac{1}{2}([0.02](1)^2 + [0.02](-1)^2 ) = 0.02 M$$

With the ionic strength, we can now calculate the activity coefficients ($\gamma_i$) of ions in solution. The activity is merely the product of the concentration of each species and the activity coefficient. $$\textrm{activity} = a_i = \gamma_i[i]$$

Let's look at a typical equilibrium expression:

$$\ce{A + B <=> C + D} $$

$$K_{eq} = \frac{[C]\gamma_c[D]\gamma_d}{[A]\gamma_a[B]\gamma_b}$$

The Extended Debye-Huckel equation is used to find the effect of ionic strength on the activity coefficent. It is given as above:

$$\log_{\gamma \ce{H+}}=\frac{-.51(z_i^2)(\sqrt{\mu})}{1+(\alpha(\frac{\sqrt{\mu}}{305})}$$

where $\alpha$ is the size of the hydrated ion in pm.

The Extended Debye-Huckel equation is valid from $\mu = 0 \ \text{to} \ 0.1 \ \text{M}.$

To find pH, you must first find $\mu$ of the hydrogen ion. You can then use the Extended Debye-Huckel equation to find the $log_{\gamma \ce{H+}}$. Take the antilog.

$$a_{\ce{H+}} = \gamma_{\ce{H+}}[\ce{H+}] $$

Take the negative log and you will get pH.

$$ pH = -\log( a_{\ce{H+}} ) = -\log( \gamma_{\ce{H+}}[\ce{H+}] ) $$

In context with the problem:

$$\log_{\gamma_{\ce{H+}}} = \frac{-.51(1^2)(\sqrt{0.02})}{1+(900 (\frac{\sqrt{0.02}}{305})}$$

$$\log_{\gamma_{\ce{H+}}} = -0.0508886 $$

$$\gamma_{\ce{H+}} = 1.1243166 $$

$$\textrm{pH} = -\log a_{\ce{H+}} = -\log (\gamma_{\ce{H+}} \cdot 1.0 \cdot 10^{-7}) = 6.95 $$

Your error is when you take the antilog of the $\log_{\gamma \ce{H+}}$. You raised it to the $10^{\log_{\gamma \ce{H+}}}$ which is $0.89429197$. It is actually supposed to be $10^{-(-0.0508886)}$ which comes out to be $1.124316$.

For the 0.10 M of $\ce{NaCl}$ I calculated the pH to be $6.92$.

  • $\begingroup$ Okay so here is where I'm having trouble, I am solving the debye-huckel equation with $u$= $0.02$ and $\alpha$ = $900$ and am getting 0.889 as my activity coefficient. I took this and multiplied it by $[H^+]$ which I'm assuming comes from $kw$=$[H^+][Oh^-]$ which is $1*10^{-7}$ multiplying that by the coefficient and then taking the $-log$ i am getting $7.05$. $\endgroup$ – John Snow Apr 1 '14 at 21:33
  • $\begingroup$ @JohnSnow I see. Of course, in a neutral solution $\ce{[H+][OH-]} = K_w$, let me use those values and see where you are going wrong. $\endgroup$ – Jun-Goo Kwak Apr 1 '14 at 21:36
  • $\begingroup$ @JohnSnow You just made a small sign error in taking the antilog. $\endgroup$ – Jun-Goo Kwak Apr 1 '14 at 21:45
  • $\begingroup$ Okay thank you for that, but now the pH is $6.95$ which is closer to the right number but we're still not arriving at $6.99$ which is the answer from the answer key. $\endgroup$ – John Snow Apr 1 '14 at 21:52
  • $\begingroup$ @JohnSnow I reviewed my calculations. I carried out the calculation and did not drop any significant figures. It is possible the answer key may have truncation errors. Did you try doing the calculation over? What value are you arriving at? $\endgroup$ – Jun-Goo Kwak Apr 1 '14 at 21:56

Using the formula as:

$$\log \gamma_\ce{H+} = \frac{- 0.51(z_i^2)\sqrt{\mu}}{1 + \sqrt{\mu}}$$

instead of your formula , this formula gives more accurate answer: $$\frac{-0.51 \times \sqrt{0.02}}{1+\sqrt{0.02}}=-0.063188664981$$ then $$10^{-(-0.063188664981)} =1.063188664981$$ then $$\mathrm{pH}=-\log(1.063188664981 \times 10^{-7})\\ \mathrm{pH}=6.973389662203$$

  • $\begingroup$ Welcome to Chemistry! Please visit this page, this page and this ‎one on how to format your posts better with MathJax and Markdown. Please type in and format the content of your image as images aren't searchable. $\endgroup$ – M.A.R. ಠ_ಠ Nov 22 '16 at 21:59
  • 1
    $\begingroup$ What does this provide that the other answer doesn’t? $\endgroup$ – Jan Nov 22 '16 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.