Part A A 110.0 mL buffer solution is 0.110 M in NHs and 0.130 M in NH4Br. What m

Question

Part A A 110.0 mL buffer solution is 0.110 M in NHs and 0.130 M in NH4Br. What mass of HCl can this buffer neutralize before the pH falls below 9.00? Express the mass in grams to three significant figures You may want to reference ( pages 781-791) Section 17.2 while completing this problem. Submit My Answers Give Up Part B and 0.395 M in NH4Br, what mass of HCI If the same volume of the buffer were 0.270 M in NHs could be handled before the pH falls below 9.00? Express the mass in grams to three significant figures. Submit My Answers Give Up

Explanation / Answer

PART A:

pKb of NH= 4.74

Initial number of moles of NH

= ([NH]) (volume of buffer solution in L)

= 0.110 M x 0.110 L

= 0.0121 moles

Initial number of moles of NH4+

= ([NH4+]) (volume of buffer solution in L)

= 0.130 M x 0.110 L

= 0.0143 moles

pOH= pKb+log([NH]/[NH])

       = 4.74 + log (0.130/0.110)

       = 4.74 + 0.07

      = 4.81

pH= 14 - 4.81 = 9.19

When the pH is 9.00, pOH = 14 – 9 = 5.00

So,

5.00 - 4.74 = log([NH]/[NH])

log([NH]/[NH]) = 0.26

[NH]/[NH]= 100.26 = 1.82

Number of moles of NH/number of moles of NH= 1.82

When HCl is added, the following reaction occurs:

NH(aq)+HCl(aq)NHCl(aq)

According to the equation, the ratio of the number of moles of NH that reacts to that of HCl that reacts is 1:1. Let x be the number of moles of HCl added.

Number of moles of NH remaining after the reaction = 0.0121 - x

Number of moles of NH present after the reaction = 0.0143 +x

So,

(0.0143 + x) / (0.0121 - x) = 1.82

0.0143 + x = 1.82 (0.0121 - x)

0.0143 + x = 0.022022 – 1.82x

2.82x = 0.022022 – 0.0143

2.82x = 0.007722

x = 0.007722 / 2.82

x = 0.00274

Molar mass of HCl = 36.5 g/mol

So, 1 mole of HCl= 36.5 g

0.00274 mole of HCl= 0.00274 x 36.5 g = 0.10001 g

                                                                 = 0.100 g correct to 3 sf

Thus, if more than 0.100 g of HCl were added to this buffer solution, its pH would fall below 9.00.

PART B:

pKb of NH= 4.74

Initial number of moles of NH

= ([NH]) (volume of buffer solution in L)

= 0.270 M x 0.110 L

= 0.0297moles

Initial number of moles of NH4+

= ([NH4+]) (volume of buffer solution in L)

= 0.395 M x 0.110 L

= 0.04345 moles

pOH= pKb+log([NH]/[NH])

       = 4.74 + log (0.130/0.110)

       = 4.74 + 0.07

      = 4.81

pH= 14 - 4.81 = 9.19

When the pH is 9.00, pOH = 14 – 9 = 5.00

So,

5.00 - 4.74 = log([NH]/[NH])

log([NH]/[NH]) = 0.26

[NH]/[NH]= 100.26 = 1.82

Number of moles of NH/number of moles of NH= 1.82

When HCl is added, the following reaction occurs:

NH(aq)+HCl(aq)NHCl(aq)

According to the equation, the ratio of the number of moles of NH that reacts to that of HCl that reacts is 1:1. Let x be the number of moles of HCl added.

Number of moles of NH remaining after the reaction = 0.0297 - x

Number of moles of NH present after the reaction = 0.04345 +x

So,

(0.04345 +x) / (0.0297 - x) = 1.82

0.04345 +x = 1.82 (0.0297 - x)

0.04345 +x = 0.054054 – 1.82x)

2.82x = 0.054054 – 0.04345

2.82x = 0.010604

x = 0.010604 / 2.82

x = 0.00376

Molar mass of HCl = 36.5 g/mol

So, 1 mole of HCl= 36.5 g

0.00376mole of HCl= 0.00376 x 36.5 g = 0.13724g

                                                                 = 0.137 correct to 3 sf

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.