5 alignment! edited Mar 14 '18 at 13:20 Gaurang Tandon 5,47988 gold badges3030 silver badges7171 bronze badges At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&(10^{-7}+10^{-2}-x)~\pu{M}&(10^{-7}-x)~\pu{M}\\ \end{array}$$$$\begin{array}{ccc} &\ce{H2O(l) &<=>& H+(aq)&+&OH-(aq)}\\\hline -&-&&\pu{10^-7M}&&\pu{10^-7M}\\ t=0&-&&\pu{(10^-7+10^-2)M}&&\pu{10^-7M}\\ t=t_0&-&&(10^{-7}+10^{-2}-x)~\pu{M}&&(10^{-7}-x)~\pu{M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&(10^{-7}+10^{-2}-x)~\pu{M}&(10^{-7}-x)~\pu{M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=>& H+(aq)&+&OH-(aq)}\\\hline -&-&&\pu{10^-7M}&&\pu{10^-7M}\\ t=0&-&&\pu{(10^-7+10^-2)M}&&\pu{10^-7M}\\ t=t_0&-&&(10^{-7}+10^{-2}-x)~\pu{M}&&(10^{-7}-x)~\pu{M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. 4 some nitpicking edit approved Mar 14 '18 at 13:17 Apoorv Potnis 71977 silver badges2626 bronze badges At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&\pu{(10^-7+10^-2-x)M}&\pu{(10^-7-x)M}\\ \end{array}$$$$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&(10^{-7}+10^{-2}-x)~\pu{M}&(10^{-7}-x)~\pu{M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&\pu{(10^-7+10^-2-x)M}&\pu{(10^-7-x)M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&(10^{-7}+10^{-2}-x)~\pu{M}&(10^{-7}-x)~\pu{M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. 3 added 241 characters in body edited Mar 14 '18 at 13:06 Gaurang Tandon 5,47988 gold badges3030 silver badges7171 bronze badges At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&\pu{(10^-7+10^-2-x)M}&\pu{(10^-7-x)M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. At a particular temperature, the $$K_{\text{eq}}$$ for the following reaction (yes, it's the auto-protolysis of water): $$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$ will be constant. Note that $$K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $$([\ce{H2O}])$$ is assumed to be constant. So, you can directly observe from here, that since $$\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$$ at a particular temperature, if $$\ce{[H+(aq)]}$$ increases (i.e. $$\mathrm{pH}$$ decreases), $$[\ce{OH-(aq)}]$$ must decrease (i.e. $$\mathrm{pOH}$$ must increase). Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $$Q$$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $$\ce{H+}$$ ions as much as it decreases the concentration of $$\ce{OH-}$$ ions. See explanation below: Consider adding $$\pu{0.01M}$$ $$\ce{HCl}$$ to pure water (at $$\pu{25^\circ C}$$) at $$t=0$$. This is what happens till achieving equilibrium at $$t=t_0$$: $$\begin{array}{ccc} &\ce{H2O(l) &<=> H+(aq)& + OH-(aq)&}\\\hline -&-&\pu{10^-7M}&\pu{10^-7M}\\ t=0&-&\pu{(10^-7+10^-2)M}&\pu{10^-7M}\\ t=t_0&-&\pu{(10^-7+10^-2-x)M}&\pu{(10^-7-x)M}\\ \end{array}$$ Now, if you actually solve for $$x$$ at $$t=t_0$$, you'll find $$x$$ to be of the order $$10^{-8}$$ or below. Notice that, for the $$\ce{H+}$$ ions, this means a decreases in concentration from $$\approx\pu{0.01M}$$ at $$t=0$$ to $$\pu{0.0099999M}$$ at $$t=t_0$$. This a hardly noticeable change. However, it is a catastrophe for the $$\ce{OH-}$$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $$\pu{10^-7 M}$$! The change in $$\mathrm{pOH}$$ is, hence, much more pronounced. 2 added 241 characters in body edited Mar 14 '18 at 12:59 Gaurang Tandon 5,47988 gold badges3030 silver badges7171 bronze badges 1 answered Mar 14 '18 at 12:53 Gaurang Tandon 5,47988 gold badges3030 silver badges7171 bronze badges