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Wize University Chemistry Textbook > Buffers and Titrations

Acid + Base Reactions

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Strong Base and Strong Acid Reactions


HCl(aq) + NaOH(aq)→ H2O(l) + NaCl(aq)HCl_{\left(aq\right)}\ +\ NaOH_{\left(aq\right)}\rightarrow\ H_2O_{\left(l\right)}\ +\ NaCl_{\left(aq\right)}

Case #1: Strong Acid is in Excess



  • We are left with moles of strong acid (HCl here) and moles of a neutral salt (NaCl here)
  • To determine the pH of the resulting solution, take the moles of HCl we are left with and use it to find moles of H+, then we need to calculate [H+]
  • n=cv so c=n/v=moles of H+/total volume of solution
  • Then use pH=-log[H+]!
The resulting solution is strongly acidic!

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Case #2: Strong Base is in Excess


  • We are left with moles of strong base (NaOH here) and moles of a neutral salt (NaCl here)
  • To determine the pH of the resulting solution, take the moles of NaOH we are left with and use it to find the moles of OH-, then we need to calculate [OH-]
  • n=cv so c=moles of OH-/total volume of solution
  • Calculate pOH using pOH=-log[OH-], and finally calculate pH using pH+pOH=14
The resulting solution is strongly basic!

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Case #3: Reacting the Same Amount of Moles of Strong Acid and Base


  • We are left with no moles of strong acid or strong base. Only the neutral salt is left
  • This means the pH=7 (neutral solution)
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Strong Base + Weak Acid Reactions


NaOH(aq)+CH3COOH(aq)→NaCH3COO(aq)+H2O(l)NaOH_{\left(aq\right)}+CH_3COOH_{\left(aq\right)}\rightarrow NaCH_3COO_{\left(aq\right)}^{ }+H_2O_{\left(l\right)}

Case #1: Strong Base is in Excess


  • We are left with strong base and a weakly basic salt
  • The resulting solution is strongly basic
  • To determine the pH use the moles of strong base to find the moles of OH-, then using n=cv, solve for the concentration of OH-
  • [OH-]=moles of OH-/total volume of solution
  • Calculate pOH using pOH=-log[OH-], then calculate pH using pH+pOH=14

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Case #2: Weak Acid is in Excess

  • The resulting solution has weak acid and a weakly basic salt
  • Note: We have a conjugate acid (CH3COOH) and its conjugate base (CH3COO-)! These are the requirements for a buffer so we have a buffer solution! We will look at buffers soon :)

CH3COOH(aq)+H2O(l) ⇌  CH3COO(aq)−+H3O(aq)+CH_3COOH\left(aq\right)+H_2O_{\left(l\right)}\ \rightleftharpoons\ \ CH_3COO_{\left(aq\right)}^-+H_3O_{\left(aq\right)}^+


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Case #3: React the Same Amount of Moles of Strong Base and Weak Acid


  • We are only left with a weakly basic salt so the solution is weakly basic.
  • To calculate pH of the resulting solution, we would need to consider the weak base:
CH3COO(aq)−+H2O(l) ⇌  CH3COOH(aq)+OH(aq)−CH_3COO_{\left(aq\right)}^-+H_2O_{\left(l\right)}\ \rightleftharpoons \ \ CH_3COOH_{\left(aq\right)}+OH_{\left(aq\right)}^-
  • After using an ICE table and finding the number of moles of OH- we end up with, we can calculate [OH-] using c=n/v=moles of OH-/total volume of solution
  • Then calculate pOH, then finally pH
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Weak Base + Strong Acid Reactions


NH3(aq)+HCl(aq)−>NH4(aq)++Cl(aq)−NH_{3\left(aq\right)}+HCl_{\left(aq\right)}->NH_{4\left(aq\right)}^++Cl_{\left(aq\right)}^-

Case #1: Strong Acid is in Excess


  • We are left with a strong acid and a weakly acidic salt
  • The resulting solution is strongly acidic
  • Determine the pH using moles of strong acid, find the [H+], find the pH

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Case #2: Weak Base is in Excess



  • We are left with a weak base and a weakly acidic salt
  • Note: We have a weak base (NH3) and its conjugate acid (NH4+) which are needed to form a buffer!
NH4(aq)++H2O(l) ⇌  NH3(aq) + H3O(aq)+NH_{4_{\left(aq\right)}}^++H_2O_{\left(l\right)}\ \rightleftharpoons\ \ NH_{3\left(aq\right)}\ +\ H_3O_{\left(aq\right)}^+


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Case #3: Reacting the Same Amount of Moles of Strong Acid and Weak Base


  • Here we are left with just the weakly acidic salt so the solution is weakly acidic.
  • Calculate pH using the weak acid:
NH4(aq)++H2O(l) ⇌  NH3(aq)+H3O(aq)+NH_{4\left(aq\right)}^++H_2O_{\left(l\right)}\ \rightleftharpoons\ \ NH_{3\left(aq\right)}+H_3O_{\left(aq\right)}^+

Practice: Calcultating the pH of the Solution When an Acid and Base React

100 mL of a 0.5 M solution of nitric acid (HNO3\rm HNO_3) was added to 75 mL of a 0.37 M solution of sodium benzoate (C6H5COONa\rm C_6H_5COONa) the conjugate base of benzoic acid (C6H5COOHKa=6.6×10−5\rm C_6H_5COOH Ka = 6.6 \times 10^{-5}). Calculate the pH\rm pH of the resulting solution.
Extra Practice
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution
Acids and Bases: Titrations
25mL of a solution of 0.5mol/L KOH is required to neutralize 15mL of sulphuric acid. What is the concentration of the acid?
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(ℓ)2KOH\left(aq\right)+H_2SO_4(aq)\to K_2SO_4(aq)+2H_2O(\ell)
100 mL of a 0.5 M solution of nitric acid (HNO3\rm HNO_3) was added to 75 mL of a 0.37 M solution of sodium benzoate (C6H5COONa\rm C_6H_5COONa) the conjugate base of benzoic acid (C6H5COOHKa=6.6×10−5\rm C_6H_5COOH Ka = 6.6 \times 10^{-5}). Calculate the pH\rm pH of the resulting solution.
Strong Acids and Bases
Calculate the pH of a solution prepared by adding 100.0 mL of 3.00 x 10-2 M KOH with 50.0 mL of 2.50 x 10-2 M Be(OH)2.
If a strong base is mixed with an equally but oppositely strong acid, what is the relative pH of the resulting salt?
Acids and Bases: Titrations
25L of a solution of 0.5M KOH is required to neutralize 15L of sulphuric acid. What is the concentration of the acid? Balance the equation if it is unbalanced.
KOH + H2SO4 --> K2SO4 + 2H2O
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution
Acids and Bases
X is a diprotic acid while Y is a triprotic acid. When X or Y are reacted with the base Ca(OH)2, which of the following statements would be true?