why a 0.010 M HCl has a pH of 2.00 but a 0.010 M HCH3COO has pH 3.37

HCl is strong acid and dissociates completely to form H+ ions and has lower pH but HCH3COO is a weak acid and so does not dissociate completely and forms less H+ ions

HCH3COO is strong acid and dissociates completely to form H+ ions and has lower pH but HCl is a weak acid and so does not dissociate completely and forms less H+ ions

smaller the ka, larger the pka, weaker the acid

smaller the ka, larger the pka, stronger the acid

smaller the kb, larger the pkb, weaker the base

smaller the kb, larger the pkb, stronger the base

([H3O+]equilibrium/[HA]initial)*100

If 2 different acids are of equal concentration, the weaker acid will have a lower percent ionization.

If two sample of the same acid are of different concentrations, the solution of lower concentration will have a higher percent ionization.

Henderson-Hasselbalch Equation

pH = pKa + 𝑙𝑜𝑔 [𝐴−]/[HA]

pOH = pKb + 𝑙𝑜𝑔 [BH+]/[B]

pH = pKa + 𝑙𝑜𝑔 [H𝐴]/[A-]

pOH = pKb + 𝑙𝑜𝑔 [B]/[BH+]

when weak acid titrated with strong base HA (aq) + OH-(aq) ⇌ A-(aq)+ H2O (l) If the weak acid is in excess,

then the resulting solution is a buffer solution. The Henderson- Hasselbalch equation can be used to determine the pH of the resulting solution

then the pH of the resulting solution will be calculated from the molarity of the solution formed from the excess moles of the base divided by the total volume of the solution

then the pH of the resulting solution will be determined by the conjugate base of the weak acid. This is the equivalence point.  Use an ICE diagram to find the equilibrium concentration of the OH-  Find the Kb (Kw/Ka) to use as your K constant.  Use the moles of the conjugate base and the total volume to find the initial molarity.  Complete the ICE diagram for the reaction shown below: A-(aq) + H2O(l) ⇌ HA(aq) + OH-(aq)

when weak acid titrated with strong base HA (aq) + OH-(aq) ⇌ A-(aq)+ H2O (l) If the strong base is in excess,

then the pH of the resulting solution will be calculated from the molarity of the solution formed from the excess moles of the base divided by the total volume of the solution

then the resulting solution is a buffer solution. The Henderson- Hasselbalch equation can be used to determine the pH of the resulting solution

then the pH of the resulting solution will be determined by the conjugate base of the weak acid. This is the equivalence point.  Use an ICE diagram to find the equilibrium concentration of the OH-  Find the Kb (Kw/Ka) to use as your K constant.  Use the moles of the conjugate base and the total volume to find the initial molarity.  Complete the ICE diagram for the reaction shown below: A-(aq) + H2O(l) ⇌ HA(aq) + OH-(aq)

when weak acid titrated with strong base HA (aq) + OH-(aq) ⇌ A-(aq)+ H2O (l) If the weak acid and strong base are equimolar

then the pH of the resulting solution will be determined by the conjugate base of the weak acid. This is the equivalence point.  Use an ICE diagram to find the equilibrium concentration of the OH-  Find the Kb (Kw/Ka) to use as your K constant.  Use the moles of the conjugate base and the total volume to find the initial molarity.  Complete the ICE diagram for the reaction shown below: A-(aq) + H2O(l) ⇌ HA(aq) + OH-(aq)

then the resulting solution is a buffer solution. The Henderson- Hasselbalch equation can be used to determine the pH of the resulting solution

then the pH of the resulting solution will be calculated from the molarity of the solution formed from the excess moles of the base divided by the total volume of the solution

When a weak base and a strong acid react B (aq) + H3O+ (aq) ⇌ HB+ (aq) + H2O (l). If the weak base is in excess,

then the resulting solution is a buffer solution and you should use the Henderson-Hasselbalch equation to determine the pH of the resulting solution.

then the pH of the resulting solution will be calculated from the molarity of the solution formed from the excess moles of the hydronium ion divided by the total volume of the solution

The pH at the equivalence point when a weak base and strong acid are reacting is determined by the conjugate acid of the weak base.  Use an ICE diagram to find the equilibrium concentration of the H3O+  Start by calculating the initial molarity of the weak conjugate acid by using the moles of the conjugate acid and the total volume.  Then complete the ICE table establishing equilibrium as shown in the reaction below. HB+(aq) + H2O(l) ⇌ B(aq) + H3O+ (aq)

When a weak base and a strong acid react B (aq) + H3O+ (aq) ⇌ HB+ (aq) + H2O (l). If the strong acid is in excess,

then the pH of the resulting solution will be calculated from the molarity of the solution formed from the excess moles of the hydronium ion divided by the total volume of the solution

then the resulting solution is a buffer solution and you should use the Henderson-Hasselbalch equation to determine the pH of the resulting solution.

The pH at the equivalence point when a weak base and strong acid are reacting is determined by the conjugate acid of the weak base.  Use an ICE diagram to find the equilibrium concentration of the H3O+  Start by calculating the initial molarity of the weak conjugate acid by using the moles of the conjugate acid and the total volume.  Then complete the ICE table establishing equilibrium as shown in the reaction below. HB+(aq) + H2O(l) ⇌ B(aq) + H3O+ (aq)

When a weak base and a strong acid react B (aq) + H3O+ (aq) ⇌ HB+ (aq) + H2O (l). If the weak base and strong acid are equimolar

The pH at the equivalence point when a weak base and strong acid are reacting is determined by the conjugate acid of the weak base.  Use an ICE diagram to find the equilibrium concentration of the H3O+  Start by calculating the initial molarity of the weak conjugate acid by using the moles of the conjugate acid and the total volume.  Then complete the ICE table establishing equilibrium as shown in the reaction below. HB+(aq) + H2O(l) ⇌ B(aq) + H3O+ (aq)

then the resulting solution is a buffer solution and you should use the Henderson-Hasselbalch equation to determine the pH of the resulting solution.

then the pH of the resulting solution will be calculated from the molarity of the solution formed from the excess moles of the hydronium ion divided by the total volume of the solution

For weak acid Vs strong base titrations

pka=pH at half equivalence point

solution is basic(pH>7) at equivalence point

solution is acidic(pH<7) at equivalence point

Where is the buffering portion on the graph?

before equivalence around half equivalence

select the correct choice for weak acid vs strong base titration

1: only HA 2:HA + A- 3:mostly A- 4:mostly OH-

1: HA + A- 2:only HA 3:mostly A- 4:mostly OH-

1: only HA 2:mostly A- 3:HA + A- 4:mostly OH-

1: only A- 2:HA + A- 3:mostly HA 4:mostly OH-

For weak acid vs Strong base:

Before half equivalence point [HA]>[A-]

After equivalence point [HA]<[A-]

Before equivalence point [HA]<[A-]

After equivalence point [HA]>[A-]

For weak acid vs Strong base:

Before half equivalence point [HA]>[A-]

After equivalence point [HA]<[A-]

Before equivalence point [HA]<[A-]

After equivalence point [HA]>[A-]

Which of the following would result in a buffer solution?

100. mL of 1.0 M CH3NH2 + 100. mL of 0.50 M CH3NH3+

100. mL of 1.0 M HNO2 + 100. mL of 2.0 M NaNO2

100. mL of 1.0 M HCl + 50.0 mL of 1.0 NaCl

100. mL of 1.0 M HF + 50.0 mL of 1.0 M HClO

For Buffer HCO2H + NaHCO2

if acid is added H3O+ + HCO2- -->HCO2H

If base is added OH- + HCO2H -->H2O + HCO2-

if acid is added H3O+ + HCO2H -->H2O+HCO2-

If base is added OH- + HCO2- -->HCO2H

Select the true statements

In buffer system moles of the base > the moles of the acid, the pH changes less when an acid is added

In buffer systems moles of acid> moles of base, pH will change less when a base is added

In buffer system moles of the base > the moles of the acid, the pH changes less when a base is added

In buffer systems moles of acid> moles of base, pH will change less when an acid is added

The buffer capacity can be increased by

Increasing the concentration of the buffer components (while keeping the ratio of these concentrations constant)

decreasing the concentration of the buffer components (while keeping the ratio of these concentrations constant)

Increasing the concentration of the acid component

Increasing the concentration of the base component

Chem Unit8

Authored by Kunal Shah

Chemistry

6th Grade

Used 2+ times

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MULTIPLE SELECT QUESTION

45 sec • 1 pt

pH

-log[H+]

14-pOH

-log[OH-]

>7 for basic solutions

>7 for acidic solutions

MULTIPLE SELECT QUESTION

45 sec • 1 pt

pOH

-log[H+]

14-pH

-log[OH-]

<7 for basic solutions

<7 for acidic solutions

MULTIPLE SELECT QUESTION

45 sec • 1 pt

pKw

1*10^-14 at 25 deg C

[H+][OH-]

>1*10^-14 @ 50 deg C

<1*10^-14 @ 50 deg C

MULTIPLE SELECT QUESTION

45 sec • 1 pt

pKw

14 at 25 deg C

-logKw

pOH + pH

pOH/pH

pH/pOH

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

pH of 0.1 M NaOH

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

pOH of 10^-3 M HCl

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

pH of 0.00345 M Sr(OH)2

0.00345

2.16

11.34

0.0069

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Chem Unit8

pH

pOH

pKw

pKw

pH of 0.1 M NaOH

pOH of 10^-3 M HCl

pH of 0.00345 M Sr(OH)2

Which has the lower pH, 0.0010 M KOH, or 0.0010 M Ca(OH)2?

which aqueous solution is a strong electrolyte?

pH of 0.005M HCl

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