Calculate the Kb for the base C2H5NH2 if the Ka for the corresponding acid is 1.8 x 10-11

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Calculate the Kb for the base C2H5NH2 if the Ka for the corresponding acid is 1.8 x 10-11 Determine the [OH-], [H3O +], and the pH of a solution that has a pOH of 9.50. The pH of 0.10 M formic acid, HCOOH, is 2.39. What is the Ka of this acid? What is the pH of a 0.40 M NH3 solution? Kb = 1.8 x 10-5. Find the pH of a 0.036 M nitrous acid (HNO2) solution. Ka = 4.5 x 10-4 HNO2 + H2O H3O+ + NO2- What is the equilibrium concentration of HNO2 and NO2- in the solution? Calculate the percent ionization for the reaction in question 12. Calculate the pH and the concentration o f all species in solution at equilibrium for a 0.15 M solution of ascorbic acid, or H2C6H6O6. For ascorbic acid, Ka1 = 8.0 10-5 and Ka2 = 1.6 10-12 Calculate the pH of a buffer that was prepared by mixing 25.0 mL of M HAc and 20.0 mL 0.150 M Ac. The Ka for the reaction is 1.8 x 10-5 What is the pH o f the buffer in question 13 after 5.0 mL of 0.075 M HCL is added to it? Calculate the pH at the following points in the titration of 50.0 mL of 0.15 M HC2H3O with 0.15 M NaOH Ka = 1.8 x 10-5 After the addition o f 25.0 mL of 0.15 M NaOH. After the addition of 50.0 mL of 0.15 M NaOH After the addition of 60.0 mL of 0.15 M NaOH.

The Correct Answer and Explanation is :

This is a multi-part question, and I’ll break it down step by step to calculate the required values. Here is the solution and explanation for each part.

1. Calculate KbK_b for C2H5NH2C_2H_5NH_2:

Given:

  • KaK_a for C2H5NH3+=1.8×10−11C_2H_5NH_3^+ = 1.8 \times 10^{-11}
  • Kw=1.0×10−14K_w = 1.0 \times 10^{-14}

Kb=KwKa=1.0×10−141.8×10−11=5.56×10−4K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-11}} = 5.56 \times 10^{-4}

2. Determine [OH⁻], [H₃O⁺], and pH if pOH=9.50pOH = 9.50:

[OH−]=10−pOH=10−9.50=3.16×10−10[OH^-] = 10^{-\text{pOH}} = 10^{-9.50} = 3.16 \times 10^{-10} [H3O+]=Kw[OH−]=1.0×10−143.16×10−10=3.16×10−5[H_3O^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{3.16 \times 10^{-10}} = 3.16 \times 10^{-5} pH=14−pOH=14−9.50=4.50\text{pH} = 14 – \text{pOH} = 14 – 9.50 = 4.50

3. Calculate KaK_a for formic acid (HCOOH):

Given:

  • [HCOOH]=0.10 M[HCOOH] = 0.10 \, \text{M}
  • pH = 2.39

[H3O+]=10−pH=10−2.39=4.07×10−3[H_3O^+] = 10^{-\text{pH}} = 10^{-2.39} = 4.07 \times 10^{-3}

Assume x=[H3O+]=[HCOO−]x = [H_3O^+] = [HCOO^-] at equilibrium: Ka=x2[HCOOH]−x≈(4.07×10−3)20.10=1.66×10−4K_a = \frac{x^2}{[HCOOH] – x} \approx \frac{(4.07 \times 10^{-3})^2}{0.10} = 1.66 \times 10^{-4}

4. Calculate pH of 0.40 M NH3NH_3:

Given Kb=1.8×10−5K_b = 1.8 \times 10^{-5}: [OH−]=Kb⋅[NH3]=(1.8×10−5)(0.40)=2.68×10−3[OH^-] = \sqrt{K_b \cdot [NH_3]} = \sqrt{(1.8 \times 10^{-5})(0.40)} = 2.68 \times 10^{-3} pOH=−log⁡[OH−]=−log⁡(2.68×10−3)=2.57\text{pOH} = -\log[OH^-] = -\log(2.68 \times 10^{-3}) = 2.57 pH=14−pOH=14−2.57=11.43\text{pH} = 14 – \text{pOH} = 14 – 2.57 = 11.43

5. pH of 0.036 M HNO2HNO_2:

Given Ka=4.5×10−4K_a = 4.5 \times 10^{-4}: [H3O+]=Ka⋅[HNO2]=(4.5×10−4)(0.036)=4.02×10−3[H_3O^+] = \sqrt{K_a \cdot [HNO_2]} = \sqrt{(4.5 \times 10^{-4})(0.036)} = 4.02 \times 10^{-3} pH=−log⁡[H3O+]=−log⁡(4.02×10−3)=2.40\text{pH} = -\log[H_3O^+] = -\log(4.02 \times 10^{-3}) = 2.40

Equilibrium Concentrations:

  • [HNO2]=0.036−[H3O+]≈0.036−0.00402=0.03198 M[HNO_2] = 0.036 – [H_3O^+] \approx 0.036 – 0.00402 = 0.03198 \, \text{M}
  • [NO2−]=[H3O+]=4.02×10−3 M[NO_2^-] = [H_3O^+] = 4.02 \times 10^{-3} \, \text{M}

Percent Ionization:

%Ionization=[H3O+][HNO2]initial×100=4.02×10−30.036×100=11.17%\% \text{Ionization} = \frac{[H_3O^+]}{[HNO_2]_{\text{initial}}} \times 100 = \frac{4.02 \times 10^{-3}}{0.036} \times 100 = 11.17\%

6. pH and species concentrations for 0.15 M ascorbic acid (H2C6H6O6H_2C_6H_6O_6):

First dissociation:

[H3O+]=Ka1⋅[H2C6H6O6]=(8.0×10−5)(0.15)=3.46×10−3[H_3O^+] = \sqrt{K_{a1} \cdot [H_2C_6H_6O_6]} = \sqrt{(8.0 \times 10^{-5})(0.15)} = 3.46 \times 10^{-3} pH=−log⁡[H3O+]=2.46\text{pH} = -\log[H_3O^+] = 2.46

Second dissociation is negligible due to Ka2K_{a2}, so:

  • [H3O+]=3.46×10−3 M[H_3O^+] = 3.46 \times 10^{-3} \, \text{M}
  • [H2C6H6O6]=0.15−3.46×10−3≈0.147 M[H_2C_6H_6O_6] = 0.15 – 3.46 \times 10^{-3} \approx 0.147 \, \text{M}
  • [HC6H6O6−]=3.46×10−3 M[HC_6H_6O_6^-] = 3.46 \times 10^{-3} \, \text{M}

7. Buffer pH (HAc and Ac⁻):

Using the Henderson-Hasselbalch equation: pH=pKa+log⁡([A−][HA])\text{pH} = \text{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right) pH=−log⁡(1.8×10−5)+log⁡((20.0 mL×0.150 M)(25.0 mL×0.100 M))=4.74+log⁡(1.2)=4.83\text{pH} = -\log(1.8 \times 10^{-5}) + \log\left(\frac{(20.0 \, \text{mL} \times 0.150 \, \text{M})}{(25.0 \, \text{mL} \times 0.100 \, \text{M})}\right) = 4.74 + \log(1.2) = 4.83

8. Buffer pH after adding HClHCl:

Moles of HCl=5.0 mL×0.075 M=0.375 mmol\text{Moles of } HCl = 5.0 \, \text{mL} \times 0.075 \, \text{M} = 0.375 \, \text{mmol} New [A−]=3.0−0.375=2.625 mmol\text{New } [A^-] = 3.0 – 0.375 = 2.625 \, \text{mmol} New [HA]=2.5+0.375=2.875 mmol\text{New } [HA] = 2.5 + 0.375 = 2.875 \, \text{mmol} pH=4.74+log⁡(2.6252.875)=4.74−0.04=4.70\text{pH} = 4.74 + \log\left(\frac{2.625}{2.875}\right) = 4.74 – 0.04 = 4.70

9. Titration pH calculations:

(a) After 25.0 mL NaOH: Half-equivalence point: pH=pKa=4.74\text{pH} = \text{p}K_a = 4.74

(b) After 50.0 mL NaOH: Equivalence point: pH=7+12(pKb−pKa)\text{pH} = 7 + \frac{1}{2}\left(\text{p}K_b – \text{p}K_a\right) Kb=KwKa=5.56×10−10K_b = \frac{K_w}{K_a} = 5.56 \times 10^{-10} pH≈8.87\text{pH} \approx 8.87

(c) After 60.0 mL NaOH: Excess OH−OH^-: [OH−]=10 mmol110 mL=9.09×10−4[OH^-] = \frac{10 \, \text{mmol}}{110 \, \text{mL}} = 9.09 \times 10^{-4} pOH=−log⁡[OH−]=3.04\text{pOH} = -\log[OH^-] = 3.04 pH=14−3.04=10.96\text{pH} = 14 – 3.04 = 10.96

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