Gen Chem II Lab 15: Titration of Acids and Bases & Lab 16: Determination of Ka

molarity

moles of solute/ liters

moles solute =

mmol x liters

mmol solute=

M x mL

equivalence point

when moles of acid and base are equal

indicator

a compound that is added to analyze which gives a visual cue (color change) when the end point is reached

at ½ equivalence point pH=

pKa

calculation of pH before the titration

Ka (lookup) =[H₃O⁺][A^-]/[HC₂H₃O₂], ICE table, plug into Ka equation, -log [H₃O⁺]= pH

calculation of pH after titration started but before equivalence point

buffer question, find limiting reagent (LxM), subtract mol of limiting reagent from products and add mol of limiting to reactants, Henderson Hasselbach (pH= pKa+log [A^-]/[HA])

calculation of pH at equivalence point

find mol of HA / OH^- & add to product, 100% of weak acid is weak base, find Kb from Ka, find [A^-] at eq (mol/ total V), use ICE and sub into weak base equilibrium (K_b = [OH^-][HC₂H₃O₂]/(C₂H₃O₂^-]) to find [OH^-], pOH= -log[OH^-], pH=14-pOH

calculation after equivalence point

strong base equation, subtract mol of limiting reagent (HA) from reactants and add to products, find [OH^-] (mol left/total V), pOH= -log[OH^-], pH= 14- pOH

how to determine concentration of unknown at equivalence point

moles of acid / volume of unknown

analyte

what were trying to analyze

titrant

a solution of one of the reactants with a known concentration. The concentration is determined by a process known as standardization

neutralization rxn

when an acid and base mix to create water and a salt

equivalence point

the point (volume) in a titration when the number of moles of analyte (acid) is stoichiometrically equal to the moles of titrant (base)

how much base should we add before the pH hits 4 ?

0.5 mL

when should we be adding 0.5 mL of titrant

before pH 4 ad after pH 11

how many mL at a timeshould we be adding between pH 4 and pH 11 ?

0.1 mL

When do we add 0.1 mL of titrant at a time?

between pH 4 and pH 11

vol of titrant to find equivalence point

mol analyte x M_analyte, mol analyte=mol titrant, mol titrant x M_titrant

pH before EP

<7

pH after EP

>7

end point

the point (volume) in titration when the indicator changes color. Ideally Ep. Provides close approximation of EP

how to standardize a solution

add specific measurement of known substance (KHP) into water to have accurate M, add indicator, add unknown (NaOH) until indicator changes color, measure volume of unknown (NaOH) added

standardization; [NaOH]

mol NaOH(=mol KHP)/ vol NaOH added (L)

what percent of the unknown was KHP

75

determining % mass: calc mass of unknown for 20 mL of titrant

M_titrant x 0.02 L x (100/75)

determining % mass: mass of KHP in unknown

L NaOH x M_NaOH x 1 mol KHP/ 1 mol NaOH x M_KHP

detrmining % mass: % KHP

mass KHP/ mass of unknown

Ka

the disassociation constant of a weak acid

strong acid

an acid that undergoes 100% ionization in water

for a strong acid, at equilibrium [HA]=

0M

for a strong acid, at equilibrium [A^-]=

1M

Ka=

[H₃O⁺][A^-]/[HA]

weak acid

partially ionizes in water to give an equilibrium mixture of the weak acid and H₃O⁺ in solution.

what is the Ka for a weak acid ?

the degree of ionization

what is the difference between a strong acid and a weak acid at the start of their titration curves?

the weak acid is higher

units of a titration curve

pH/volume (mL)

-log₁₀[H₃O⁺]=

pH

-log₁₀[K_a]=

pKa

pH(substituted)=

pK_a+log₁₀[A^-]/[HA]

titration curve

the trend in pH of a solution of a weak or strong acid/base (analyze) as a function of the volume of strong acid/base (titrant) added.

what volume is required to reach the equivalence point (calculation)?

L_added at equivalence point x M_acid= mol acid=mol base, mol base x M_base=L

ph of STRONG acid at equivalence point

pH= 7

pH of WEAK acid at equivalence point

>7

at what point of the titration are acids neutralized

the equivalence point

how to find Kb from Ka

Kw(1×10^-14)/Ka

where do we find the equivalence point in the second derivative?

where the function crosses the zero

[HA] calculation

V_EP x M_NaOH x (1mol HA/1molNaOH) x 1/ L_HA