Question

1. An aqueous solution is prepared such that pH = 8.2. Determine the [H+] [OH–], and pOH of this solution. Show work.

2. Below are the relative fraction of species versus pH curves for different acids. Based on these curves, rank the acids (carbonic acid (H2CO3), bicarbonate (HCO3 –), phenol (C6H5OH), hydrofluoric acid (HF)) in order of increasing acid strength. Briefly describe reasoning.

3. According to the list of bases in order of increasing base strength up the page, which side of the acid-base reaction shown below is favored at equilibrium? Briefly explain. HPO4 2– (aq) + HS– (aq) H2PO4 – (aq) + S2– (aq) Base: Strongest OH– S2– PO4 3– HPO4 2– HS– H2PO4 – H2O Weakest

4. Below are the chemical structures of three species labeled A, B, and C. One is an acid, one is a base, and one is neither an acid nor a base when dissolved in water. Assign which one is which and briefly explain assignments.

Answer 1

1. Given that: pH = 8.2

We know that pH + pOH =14

Hence, pOH = 14 - pH =14 - 8.2 =5.8

As, pH = -log[H+] & pOH = - log [OH]

=> -8.2 = log [H+] & -5.8 = log [OH]

=> 10^-8.2 = [H+]. & 10^-5.8 = [OH]

=> [H+] = 6.31×10^-9 & [OH] = 1.58×10^-6 (Answer)

2. Increasing acidic order:

HCO³^- < C6H5OH < H2CO3 < HF

Explanation: Acidic strength ∝ stability of conjugate base ∝ Ka ∝ 1/pKa value.

Hence the stability of conjugate base can determine the acidic strength.

3. The equilibrium will shif to right side because more stable H2PO4 - is formed from less stable HPO4 2- as it has dianion on central P atom.

HPO4 2- (aq) + HS- (aq) -> H2PO4 -(aq) + S2- (aq)

4. When an acid dissolve in water it gives H+ ion, and when base is dissolved it gives OH- ion. If the dissolved on does not give H+ or OH- rather than it give other cation and anion then it is considered as salt. So among A, B, C which will give H+ ion is acidic in nature and which one will give OH- ion is basic in nature and which will give other cation and anion rather than H+&OH- is salt.