# calculating ph of a mixture of acids

I came across a question that couldn't find the answer to and i couldn't find the answer anywhere. The question is the following.

You have a soltion created by dissolving HCN and HBrO in water (T = 298.15K) and the pH = 4.816. A titration of 100ml of this solution needs 100ml of 0.2M NaOH to reach its point of equivalence. Wich of the following statements is true:

-the concentration of HBrO equals 1.5 * 10^-1 M in the original solution -the concentration of HCN equals 5.4 * 10^-2 M in the original solution -the concentration of BrO- equals 1.5 * 10^-1 M in the titration solution -the concentration of CN- equals 6.1 * 10^-1 M in the titration solution

I already found the following things: the concentration of H3O+ in the original solution should be 10^-4.816. I can fill this in in the acidity constants of the reaction to find:

°the concentration of CN- = 6.2*10^-10 / (10^-4.862) * the concentration of HCN °the concentration of BrO- = 2.8*10^-9 / (10^-4.862) * the concentration of HBrO

and the change of the sum of these two concentrations should equal the concentration of H3O+.

This is the point where i get stuck. I don't really know how I can use the titration to benefit my calculations. It would be a huge help if someone would help me.

I also apologise if I sometimes use wrong/weird terms to describe things. English isn't my mother tongue and I haven't got the slightest clue what the english terms are.

This proved to be a most interesting problem as posed. To solve it we note that we have two things we need to find: the amount of $\ce{HCN}$ and the amount of $\ce{HBrO}$ that went into the original solution. We also have two observations: the pH of the original solution and the amount of $\ce{NaOH}$ that was required to bring the solution to the equivalence point. Given that we have two equations in two unknowns we can get a solution if the two equations are consistent (we can find the point of intersection of crossing straight lines given the slope and intercept of each) but if they are not we cannot (we can't find the point of intersection of two parallel lines). The data here are inconsistent so there is no solution. If the strength of the NaOH solution used for the titration were 0.1 M instead of 0.2 M or if it took 50 mL of 0.2 N instead of the specified 100 to reach equivalence then we can obtain an answer which is that the original solution is 0.117 M in HCN and .081 M in HBrO. To check on this answer we note that the fraction of molecules of a monoprotic acid in a solution at pH that lose their proton is $$f_1(pH, pK_a) = 1 - 1/(1 + 10^{(pH - pK_a)})$$ This comes out of the Henderson Hasselbalch equation. Thus if we have $M_{HCN}$ molar HCN and $M_{HBrO}$ molar HBrO the number of protons released to a liter of solution at pH 4.186 would be $$N_{H^+} = M_{HCN}f_1(4.816,9.21) + M_{HBrO}f_1(4.816,8.70)$$ with 9.21 and 8.70 being, respectively, the $pK_a$ values for prussic and hypobromous acids. This is the first equation. It is given that $N_{H^+} = 10^{-4.816}$ and if you substitute $M_{HCN} = 0.117$ and $M_{HBrO} = 0.081$ into it you will see that these values satisfy it.

The second equation deals with the titration data. If we have $M_{HCN}$ molar prussic acid at some pH, here pH0 = 4.816, there are, as we have seen, $M_{HCN}f_1(pH0,9.21)$ moles of dissociated acid molecules in a liter at that pH. At some other pH, call it $pH_i$ the number of dissociated molecules HCN per liter will be $M_{HCN}f_1(pH_i,9.21)$. To get from pH0 to $pH_i$ thus requires that $$dQ_{HCN} = M_{HCN}f_1(pH_i,9.21) - M_{HCN}f_1(pH0,9.21)$$ moles of protons be absorbed by $\ce{OH^-}$ ions. There is a similar requirement for the other acid. The water itself goes from a concentration of hydrogen ions of $10^{-pH0}$ to $10^{-pH_i}$ requiring $10^{-pH0}- 10^{-pH_i}$ additional $\ce{OH^-}$ ions to neutralize them and finally, the $\ce{OH^-}$ ion concentration rises from $10^{(pH0 - 14)}$ to $10^{(pH_i - 14)}$ requiring that $10^{(pH_i - 14)} - 10^{(pH0 - 14)}$ additional. Defining the water's $\ce{OH^-}$ requirement as

$$dQ_w =10^{-pH0} - 10^{-pH_i} + 10^{(pH_i - 14)} - 10^{(pH0 - 14)}$$

we can now write the second equation $$N_{OH^-} = M_{HCN}dQ_{HCN} + M_{HBrO}dQ_{HBrO} + dQ_w$$

At this point we have our second equation but there is a big problem here. We don't know what $pH_i$ is and thus cannot obtain values for $N_{OH^-}$ to compare to the amount of base in the titrant we used. Either the question has been posed with $pH_i$ omitted in error or the poser really wants us to think. We are given the amount of base required to reach the 'equivalence point' and while it is pretty clear what 'equivalence point' means when a single acid (each acid has its equivalent point) but what's the equivalent point of a mix? If the acids are close in strength, as these two are, the composite titration curve exhibits a clear inflection point and I am interpreting equivalence point here to mean inflection point. Fortunately, the inflection point of the titration curve of a mix of acids of similar strength depends can be accurately found from log of the ratio of the strengths as long as -2 < log ratio < 2. Using this we can eliminate the functional dependence of the second equation on $pH_i$. Thus, while the second equation appears to depend on $M_{HCN}$,$M_{HBrO}$ and $pH_i$ in fact the latter is completely determined from the former 2 and so we have a second equation dependent only on $M_{HCN}$ and $M_{HBrO}$ which, along with the first (also a function of these two variables) gives us a system that can be solved as long as we are given reasonable data to plug into the system. 100 mL of 0.2 M NaOH is not reasonable data. 50 mL of 0.2 M is reasonable data. If the titration NaOH was 50 mL we would, as stated earlier, obtain $M_{HCN} = 0.117$ and $M_{HBrO} = 0.081$ from which we can calculate $pH_i = 9.20$ So what would happen if we added 100 mL of 0.2 NaOH? The mixture would come to some pH other than the inflection point pH. If we are told what that is we could still solve the problem and much more easily.

I'm not going to say much about how one solves this pair of equations as, while the first is linear in the unknowns, the second isn't because it is nonlinearly dependent on $pH_i$ which is, in turn, dependent on the unknowns is a non linear fashion. The 4 derivatives of $N_{H^+}$ and $N_{OH^-}$ are found with respect to the two unknowns are found and arranged in a 2 x 2 matrix. The is inverted and multiplied by the error vector (the difference between what values of $N_{H^+}$ and $N_{OH^-}$ we get using our guess as to what $M_{HCN}$ and \$M_{HBrO}} might be) to give a correction for our guess. This is repeated until the errors are very small. It's Newton's method for more than one unknown.

This has been, it seems, more about math than chemistry by I don't know how else one might solve this problem.