2 slight clean up mathjax edited Mar 1 '18 at 2:16 Gaurang Tandon 5,43288 gold badges3030 silver badges7070 bronze badges The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid. You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps. (For the sake of simplicity I write $$c(H^+)=x$$$$c(\ce{H+})=x$$ and $$k_{a,n}=k_n$$.) Exact: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$$$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$ Neglecting $$k_3$$: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$$$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$ Neglecting $$k_2$$ and $$k_3$$: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{k_1}{x+k_1}$$$$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{k_1}{x+k_1}$$ Solving those equations give the following pH values (with super unrealistic high precision): Exact: 1.2469291050971560 w/o $$k_3$$: 1.2469291065371300 w/o $$k_2$$ and $$k_3$$: 1.2470654318802320 As a pH can only be measured precisely within sthsomething like $$\pm \left(10^{-2}\ldots10^{-3}\right)~\text{pH}$$$$\pm \left(10^{-2}\ldots10^{-3}\right)~\mathrm{pH}$$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected. So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days. The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid. You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps. (For the sake of simplicity I write $$c(H^+)=x$$ and $$k_{a,n}=k_n$$.) Exact: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$ Neglecting $$k_3$$: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$ Neglecting $$k_2$$ and $$k_3$$: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{k_1}{x+k_1}$$ Solving those equations give the following pH values (with super unrealistic high precision): Exact: 1.2469291050971560 w/o $$k_3$$: 1.2469291065371300 w/o $$k_2$$ and $$k_3$$: 1.2470654318802320 As a pH can only be measured precisely within sth like $$\pm \left(10^{-2}\ldots10^{-3}\right)~\text{pH}$$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected. So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days. The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid. You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps. (For the sake of simplicity I write $$c(\ce{H+})=x$$ and $$k_{a,n}=k_n$$.) Exact: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$ Neglecting $$k_3$$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$ Neglecting $$k_2$$ and $$k_3$$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{k_1}{x+k_1}$$ Solving those equations give the following pH values (with super unrealistic high precision): Exact: 1.2469291050971560 w/o $$k_3$$: 1.2469291065371300 w/o $$k_2$$ and $$k_3$$: 1.2470654318802320 As a pH can only be measured precisely within something like $$\pm \left(10^{-2}\ldots10^{-3}\right)~\mathrm{pH}$$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected. So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days. 1 answered Dec 7 '15 at 14:30 pH13 - Yet another Philipp 6,67911 gold badge3030 silver badges6262 bronze badges The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid. You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps. (For the sake of simplicity I write $$c(H^+)=x$$ and $$k_{a,n}=k_n$$.) Exact: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$ Neglecting $$k_3$$: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$ Neglecting $$k_2$$ and $$k_3$$: $$x=\frac{k_W}{x}+c_0(H_3A)\frac{k_1}{x+k_1}$$ Solving those equations give the following pH values (with super unrealistic high precision): Exact: 1.2469291050971560 w/o $$k_3$$: 1.2469291065371300 w/o $$k_2$$ and $$k_3$$: 1.2470654318802320 As a pH can only be measured precisely within sth like $$\pm \left(10^{-2}\ldots10^{-3}\right)~\text{pH}$$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected. So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days.