A buffer solution is made by dissolving H₃PO₄ and NaH₂PO₄ in wate...

Question

A buffer solution is made by dissolving H₃PO₄ and NaH₂PO₄ in water. (10.8)b. Write an equation that shows how this buffer neutralizes added base.

Accepted Answer

Welcome back, everyone. We need to write a reaction equation showing how a buffer solution made from combining aqueous solutions of hydrogen cyanide and sodium cyanide will neutralize an added base. Let's begin by categorizing our two aqueous solutions. Beginning with hydrogen cyanide, we want to recognize that H C N or hydrogen cyanide is considered a weak acid since it's not present on our memorized list of strong acids in our textbook. And next, we're going to recognize that we have sodium cyanide which consists of sodium a metal and our poly atomic anion cyanide C N one minus which is again a poly atomic anion. And this is going to be considered an ionic compound. Since we have again, the combination of a metal plus poly atomic ion recall that ionic compounds will dissociate. Where in contrast, our weak acid being a weak acid is a weak electrolyte and will only undergo partial dissociation. So we'll say it partially dissociates. Now showing our dissociation of sodium cyanide, we would form N AC N which dissociates fully into our sodium plus one ion, which we should recognize as a spectator and does not influence Ph. You can also recognize this as a counter ion. And our second ion that sodium cyanide dissociates into would be our cyanide C N one minus anion in which bla based on this one minus charge is going to be considered a base since it can act as an electron pair donor or accept a proton. So in this case, we would label it as our conjugate base of our base sodium cyanide. So we're going to label this cyanide anion as a weak base. Now, going back to our prompt, we're told that our solutions will neutralize an added base. So adding a base would mean adding our hydroxide anion and recall that our hydroxide anion will want to react with our acidic component of our solution in our buffer system. So in this case, our hydroxide will not react with sodium cyanide because there is no acidic component that it contains, it consists of dissociating into a spectator and a weak base. However, our hydroxide will react with our hydrogen cyanide solution because we do have in the partial dissociation of hydrogen cyanide, which is an equilibrium on our product side, we have an acidic proton as well as our cyanide anion. So we have the presence of this acidic proton which can react with hydroxide. So writing out our reaction, we will have hydrogen cyanide which is our acid reacting with hydroxide. Recall that our acid is going to be a proton donor. So we would donate a proton to form a bond using the extra electrons in our hydroxide due to that minus one charge. And as a result, we would form our conjugate base, which is our cyanide one minus anion. And our second product is going to be water note that as we stated, hydrogen cyanide is a weak acid and it's going to be in solution with its conjugate base. And we should recall that a buffer solution can either consist of a weak acid and its conjugate base or a weak base and its conjugate acid in solution. So therefore, we do have a buffer and being that our atoms are balanced on both sides of our reaction. Our reaction as written is already bounced. We do not need to add any coefficients. And so our final answer is going to be our reaction between one mole of hydrogen cyanide reacting with one mole of hydroxide to form one mole of cyanide anion and one mole of water as products. This is our final answer. I hope this made sense and let us know if you have any questions.

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A buffer solution is made by dissolving H₃PO₄ and NaH₂PO₄ in water. (10.8)b. Write an equation that shows how this buffer neutralizes added base.

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