Question

a) Buffer solutions with a pH values of around 10 are prepared using sodium carbonate (Na2CO3) and sodium hydrogen carbonate (NaHCO3). What is the pH of a solution of 10.0g each of the two salts in enough water to make 0.250 L of solution? pKa2= 10.33

b) By how much does the pH change when 3.5mL of 6.0 M HCl is added?

c) By how much does the pH of the solution change when 0.92 g of NaOH is added?

Answer 1

Answer:-

(a)-

Given:-

wt. of sodium carbonate (Na₂CO₃) = 10.0 g

wt. of sodium hydrogen carbonate (NaHCO₃) = 10.0 g

volume of buffer solution (V) =  0.250 L

pK_a2 value = 10.33

As we know that

molar mass of sodium carbonate (Na₂CO₃) = 2 molar mass of Na + molar mass of C + 3 molar mass of O

molar mass of sodium carbonate (Na₂CO₃) = 2 23 + 12 + 3 16

molar mass of sodium carbonate (Na₂CO₃) = 46 + 12 + 48

molar mass of sodium carbonate (Na₂CO₃) = 106 g / mol

similarly

molar mass of sodium hydrogen carbonate (NaHCO₃) = molar mass of Na + molar mass of H + molar mass of C + 3 molar mass of O

molar mass of sodium hydrogen carbonate (NaHCO₃) = 23 + 1 + 12 + 3 16

molar mass of sodium hydrogen carbonate (NaHCO₃) = 23 + 1 + 12 + 48

molar mass of sodium hydrogen carbonate (NaHCO₃) = 84 g / mol

therefore

No. of moles of sodium carbonate (Na₂CO₃) = wt. of sodium carbonate (Na₂CO₃) / molar mass of sodium carbonate (Na₂CO₃)

No. of moles of sodium carbonate (Na₂CO₃) = 10.0 g / 106 g / mol

No. of moles of sodium carbonate (Na₂CO₃) = 0.094 mol

similarly

No. of moles of sodium hydrogen carbonate (NaHCO₃) = wt. of ssodium hydrogen carbonate (NaHCO₃) / molar mass of sodium hydrogen carbonate (NaHCO₃)

No. of moles of sodium hydrogen carbonate (NaHCO₃) = 10.0 g / 84 g / mol

No. of moles of sodium hydrogen carbonate (NaHCO₃) = 0.119 mol

since we know that

molar concentration of compound = No. of moles of compound (n) / volume of solution (V)

molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = No. of moles of sodium carbonate (Na₂CO₃) / volume of buffer solution (V)

molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = 0.094 mol / 0.250 L

molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = 0.376 mol /  L = 0.376 M

similarly

molar concentration of sodium hydrogen carbonate (NaHCO₃) i.e [NaHCO₃] = No. of moles of sodium hydrogen carbonate (NaHCO₃) / volume of buffer solution (V)

molar concentration of sodium hydrogen carbonate (NaHCO₃) i.e [NaHCO₃] = 0.119 mol / 0.250 L

molar concentration of sodium hydrogen carbonate (NaHCO₃) i.e [NaHCO₃] = 0.476 mol /L = 0.476 M

So according to the Henderson-Hasselbalch equation

pH = pK_a + log [salt] /[acid]

since sodium hydrogen carbonate (NaHCO₃) = acidic salt i.e acid

sodium carbonate (Na₂CO₃) = salt

therefore

pH = pK_a2 + log [Na₂CO₃] /[NaHCO₃]

pH =10.33 + log (0.376) / (0.476)

pH =10.33 + log (0.376 / 0.476)

pH =10.33 + log (0.7899)

pH =10.33 + ( - 0.1024)

pH =10.33 - 0.1024

pH =10.23 (i.e the answer)

(b)-

Given:-

volume of HCl = 3.5 mL = 3.5 / 1000 = 3.5 10^-3 L

molarity of HCl = 6.0 M = 6.0 mol / L

As we know that

molarity of compound = no. of moles of compound (n) / volume of solution (V) in liter (L)

molarity of HCl = no. of moles of HCl (n) / volume of solution (V) in liter (L)

no. of moles of HCl (n) = molarity of HCl volume of solution (V) in liter (L)

no. of moles of HCl (n) = 6.0 mol / L  ​​​​​​​ 3.5 10^-3 L

no. of moles of HCl (n) = 21 10^-3 mol

no. of moles of HCl (n) = 0.021 mol

Since we know that

No. of moles of sodium carbonate (Na₂CO₃) = 0.094 mol

No. of moles of sodium hydrogen carbonate (NaHCO₃) = 0.119 mol

volume of buffer solution (V) =  0.250 L

when addition of 3.5 mL of 6.0 M HCl i.e 0.021 mol of HCl takes place buffer solution of sodium hydrogen carbonate (NaHCO₃) and sodium carbonate (Na₂CO₃) then no. of moles of sodium carbonate (Na₂CO₃) is decreased due to reaction between sodium carbonate (Na₂CO₃) and HCl and no. of moles of sodium hydrogen carbonate (NaHCO₃) is increased as follows:-

Na₂CO₃ + HCl NaHCO₃ + NaCl

Initial 0.094 mol 0    0.119 mol 0

Change - 0.021 mol - 0.021 mol + 0.021 mol

therefore

decreased no. of moles of sodium carbonate (Na₂CO₃) = 0.094 mol - 0.021 mol = 0.073 mol

increased  no. of moles of sodium hydrogen carbonate (NaHCO₃) = 0.119 mol + 0.021 mol = 0.140 mol

So

New molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = No. of moles of sodium carbonate (Na₂CO₃) / volume of buffer solution (V)

New molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = 0.073 mol / 0.250 L

New molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = 0.292 mol /  L = 0.292 M

similarly

New molar concentration of sodium hydrogen carbonate (NaHCO₃) i.e [NaHCO₃] =0.140 mol /0.250 L

New molar concentration of sodium hydrogen carbonate (NaHCO₃) i.e [NaHCO₃] = 0.560 mol /  L = 0.560 M

So according to the Henderson-Hasselbalch equation

pH = pK_a + log [salt] /[acid]

pH = pK_a2 + log [Na₂CO₃] /[NaHCO₃]

pH =10.33 + log (0.292) / (0.560)

pH =10.33 + log (0.292 / 0.560)

pH =10.33 + log (0.5214)

pH =10.33 + ( - 0.2828)

pH =10.33 - 0.2828

pH =10.05

Change in pH of buffer solution = pH of buffer solution after addition of HCl - Initial pH of buffer solution

Change in pH of buffer solution = 10.05 - 10.23

Change in pH of buffer solution = - 0.18 pH decreased (i.e the answer)

(c)-

Given:-

wt. of NaOH  = 0.92 g

As we know that

molar mass of NaOH =   molar mass of Na + molar mass of O + molar mass of H

molar mass of NaOH =   23 + 16 + 1

molar mass of NaOH =   40 g / mol

Also we know that

No. of moles of NaOH = wt. of NaOH  / molar mass of NaOH

No. of moles of NaOH = 0.92 g  / 40 g / mol

No. of moles of NaOH = 0.023 mol

Since we know that

No. of moles of sodium carbonate (Na₂CO₃) = 0.094 mol

No. of moles of sodium hydrogen carbonate (NaHCO₃) = 0.119 mol

volume of buffer solution (V) =  0.250 L

when addition of 0.92 g of NaOH i.e 0.023 mol of NaOH takes place buffer solution of sodium hydrogen carbonate (NaHCO₃) and sodium carbonate (Na₂CO₃) then no. of moles of sodium hydrogen carbonate (NaHCO₃) is decreased due to reaction between sodium hydrogen carbonate (NaHCO₃) and NaOH and no. of moles of sodium carbonate (Na₂CO₃) is increased as follows:-

NaHCO₃ + NaOH Na₂CO₃ + H₂O

Initial 0.119 mol 0    0.094 mol 0

Change - 0.023 mol - 0.023 mol + 0.023 mol

therefore

increased no. of moles of sodium carbonate (Na₂CO₃) = 0.094 mol + 0.023 mol = 0.117 mol

decreased  no. of moles of sodium hydrogen carbonate (NaHCO₃) = 0.119 mol - 0.023 mol= 0.096 mol

So

New molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = No. of moles of sodium carbonate (Na₂CO₃) / volume of buffer solution (V)

New molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = 0.117 mol / 0.250 L

New molar concentration of sodium carbonate (Na₂CO₃) i.e [Na₂CO₃] = 0.468 mol /  L = 0.468 M

similarly

New molar concentration of sodium hydrogen carbonate (NaHCO₃) i.e [NaHCO₃] =0.096 mol /0.250 L

New molar concentration of sodium hydrogen carbonate (NaHCO₃) i.e [NaHCO₃] = 0.384 mol /  L = 0.384 M

So according to the Henderson-Hasselbalch equation

pH = pK_a + log [salt] /[acid]

pH = pK_a2 + log [Na₂CO₃] /[NaHCO₃]

pH =10.33 + log (0.468) / (0.384)

pH =10.33 + log (0.468 / 0.384)

pH =10.33 + log (1.21875)

pH =10.33 + ( 0.08591)

pH =10.33 + 0.08591

pH =10.42

Change in pH of buffer solution = pH of buffer solution after addition of NaOH - Initial pH of buffer solution

Change in pH of buffer solution = 10.42 - 10.23

Change in pH of buffer solution = 0.19 pH increased (i.e the answer)