Question

# 1) A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO. Part...

1) A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO.
Part A: What is the initial pH of this solution?
Part B: What is the pH after addition of 150.0 mg of HBr?
Part C: What is the pH after addition of 85.0 mg of NaOH?

2) For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.
Part A: 280.0 mL of pure water
Part B: 280.0 mL of a buffer solution that is 0.195 M in HCHO2 and 0.280 M in KCHO2
Part C: 280.0 mL of a buffer solution that is 0.270 M in CH3CH2NH2 and 0.250 M in CH3CH2NH3Cl

#1

We have a buffer solution made of 0.175 M HClO and 0.150 M NaClO

Part A :

pH of buffer solution can be calculated using Henderson- Hasselbalch equation, which is given below

pH = pKa + log [ Base]/[Acid]

We have [acid] = [ HClO] = 0.175 M

[Base] = [NaClO] = 0.150 M

Ka of HClO is 3.5 x 10^-8 ( Note : Ka value is taken from online resource. Please check the value given to you and make the changes accordingly)

pKa = - log ( Ka)

pKa = - log ( 3.5 x 10^-8)

pKa = 7.46

pH = 7.46 + log ( 0.150/0.175)

pH = 7.46 + ( -0.0669)

pH = 7.39 ( Note : This value of pH would vary with Ka. So please check the Ka value given to you)

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Part B

When we add HBr to this buffer, basic component of buffer which is NaClO would try to neutralize effect of acid HBr.

Let's write the reaction that takes place here

HBr + NaClO -------------> HClO + NaBr

Let's find moles of HBr , HClO and NaClO we have

mol HBr = 150 mg HBr * 1 g/1000 mg * 1mol / 80.9 g = 0.00185 mol

mol NaClO = 100 mL * 1 L/1000 mL * 0.150 mol/L = 0.015 mol

mol HClO = 100 mL * 1 L/1000 mL * 0.175 mol/L = 0.0175 mol

Let's draw an ICE table for the above equilbrium reaction

 HBr NaClO HClO NaBr I 0.00185 0.015 0.0175 0 C -0.00185 -0.00185 +0.00185 +0.00185 E 0 0.01315 0.01935 0.00185

At equilibrium we have moles of acid, HClO = 0.01935

mol of base, NaClO = 0.01315

Let's plug these values in Henderson equation

pH = pKa + log [ base/acid]

pH = 7.46 + log ( 0.01315/0.01935) ( Note : we can use mole values instead of concentration terms in this case)

pH = 7.46 + (-0.1678)

pH = 7.29

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Part C

When NaOH is added to the buffer, acidic component , HClO of the buffer tries to neutralize its effect.

The reaction given below

NaOH + HClO -----> NaClO + H2O

mol NaOH = 85 mg NaOH * 1 g/1000 g * 1 mol/40 g = 0.002125 mol

Let's draw ICE table

 HClO NaOH NaClO H2O I 0.0175 0.002125 0.015 0 C -0.002125 -0.002125 +0.002125 +0.002125 E 0.0154 0 0.0171 0.002125

Equilibrium mol of base NaClO = 0.0171 and acid , HClO= 0.0154

pH = 7.46 + log ( 0.0171/0.0154)

pH = 7.46 + 0.0455

pH = 7.51

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#2

We are adding 0.01 mol of NaOH in 280 mL of pure water.

Let's find concentration of NaOH

0.01 mol * 1/ 280 mL * 1000 mL / 1 L = 0.0357 M

[NAOH] = [OH-] = 0.0357 M

pOH = - log (0.0357)

pOH = 1.447

pH + pOH = 14

pH = 14 - 1.447 = 12.55

pH = 12.55

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Part B

We are adding 0.01 mol NaOH to a buffer solution. Here acidic component of buffer, which is HCHO2 would try to neutralize the effect of added NaOH

Let's write the reaction equation

NaOH + HCHO2 ------------> NaCHO2 + H2O

Let's calculate initial moles of acidic component HCHO2 & basic component KCHO2

mol HCHO2 = 280 mL * 1 L/1000 mL * 0.195 mol/L = 0.0546 mol

mol KCHO2 = 280 mL * 1 l/1000 mL * 0.280 mol/L = 0.0784 mol

Let's draw ICE table to see how many moles are left at equivalence point

Net ionic equation is OH- + HCHO2 ------> H2O + CHO2-

 OH- HCHO2 H2O CHO2^- I 0.001 0.0546 0 0.0784 C -0.01 -0.01 +0.01 +0.01 E 0 0.0446 0.01 0.0884

Let us use henderson equation to find pH of this buffer

pH = pKa + log (base)/(acid)

Ka of HCHO2 is 1.8 x 10^-4 ( Note : Please check the Ka value given to you and make the changes accordingly)

pKa = - log Ka

pKa = - log ( 1.8 x 10^-4)

pKa = 3.74

Substituting the values,we get

pH = 3.74 + log ( 0.0884/0.0446)

pH = 4.037

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Part C

We are adding NaOH to a buffer, therefore acidic component which is CH3CH2NH3Cl would try to neutralize the effect of added NaOH.

Let's write net ionic equation

NaOH + CH3CH2NH3Cl --------> NaCl + H2O + CH3CH2NH2

OH- + CH3CH2NH3+ -----------> CH3CH2NH2 + H2O

Let's calculate initial moles of CH3CH2NH3+ and CH3CH2NH2 we have

mol CH3CH2NH3+ =280 mL * 1L/1000 mL * 0.250 mol/L = 0.07

mol CH3CH2NH2 = 280 mL * 1L/1000 mL * 0.270 mol/L = 0.0756

Let us draw ICE table

 CH3CH2NH3+ OH- CH3CH2NH2 H2O I 0.07 0.01 0.0756 0 C -0.01 -0.01 +0.01 +0.01 E 0.06 0 0.0856 0.01

Using Henderson equation, we calculate pH as

pH = pKa + log ( base/acid)

Kb of CH3CH2NH2 is 5.6 x 10^-4 ( Note : Please check the value given to you and make the changes accordingly)

Ka = Kw/Kb

Ka = 1 x 10^-14/ 5.6 x 10^-4

Ka = 1.79 x 10^-11

pKa = -log Ka

pKa = 10.75

pH = 10.75 + log ( 0.0856/0.06)

pH = 10.9

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