How do I solve for this **Determining pH During the Titration of Dimethylamine by HBr**In this exercise, you are asked to determine the pH at various points during the titration of 39.7 mL of 0.222 M dimethylamine, \((\text{CH}_3)_2\text{NH}\), with a base ionization constant \(K_b\) of \(5.9 \times 10^{-4}\), by 0.222 M hydrobromic acid (HBr). The titration is conducted at 25 °C. State symbols are not shown for species in this problem.Calculate the pH at the following stages:(a) Before the addition of any HBr: [__________](b) After the addition of 15.9 mL of HBr: [__________](c) At the titration midpoint: [__________](d) At the equivalence point: [__________](e) After adding 61.5 mL of HBr: [__________]

Since we only answer up to 3 subparts, we'll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you'd like answered. To solve for the pH before addition of HBr, solve for the concentration of OH- using the equilibrium constant (Kb): (CH3)2NH + H2O ↔ (CH3)2NH2++ OH- Kb = 5.9 x 10-4 Now relation for Kb for equilibrium constant may be written as: Kb = CH32NH2+OH-CH32NH = 5.9 x 10-4where Kb = equilibrium constant To solve for the pH after addition of HBr, use the Henderson-Hasselbach equation. pOH = pKb + log(CH3)2NH2+(CH3)2NH pH = 14 - pOH To solve for the pH during mid-titration: pOH = pKb pH = 14 - pOH a. Before addition of HBr (CH3)2NH+H2O↔(CH3)2NH2++OH-Kb=5.9×10-4 Now, draw an ICE table: (CH3)2NH (CH3)2NH2+ OH- Initial 0.222 0 0 Change -x +x +x Equilibrium 0.222-x x x Now, solve for equilibrium constant Kb = CH32NH2+OH-CH32NH = 5.9 x 10-4xx0.222-x= 5.9 x 10-4x2= 5.9 x 10-40.222-xx2+ 5.9 x 10-4x-1.3098×10-4=0x=0.0115OH-=0.0115 MpOH=1.953pH=14-pOH=12.047b. After the addition of 15.9 ml HBr mol of HBr=0.222M0.0159L=3.5298×10-3molmol of CH32NH reacted =3.5298×10-3molmol ofCH32NH initial=0.2220.0397L=8.8134×10-3molmol ofCH32NH remaining=8.8134×10-3mol-3.5298×10-3mol=5.2836×10-3molmol ofCH32NH2+ formed=3.5298×10-3mol The chemical equation may be written as: (CH3)2NH + HBr → (CH3)2NH2++ Br- Now solve for the pOH using the Henderson Hsselbalch equation as: pOH=pKb+logCH32NH2+CH32NHpOH=-log5.9×10-4+log3.5298×10-35.2836×10-3=3.05pH=10.95c. At titration midpoint CH32NH=CH32NH2+pOH=pKb+log1pOH=pKbpOH=-log(5.9×10-4)=3.23pH=14-pOH=10.77Answer is: a. Before addition of any HBr=12.047 b. After addition of 15.9 mL HBr = 10.95 c. At titration midpoint =10.77

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Determine the pH during the titration of 39.7 mL of 0.222 M dimethylamine (CH)NH, K, = 5.9×10) by 0.222 M HBr at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem. (a) Before the addition of any HBr (b) After the addition of 15.9 mL of HBr (c) At the titration midpoint

Determine the pH during the titration of 39.7 mL of 0.222 M dimethylamine (CH)NH, K, = 5.9×10) by 0.222 M HBr at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem. (a) Before the addition of any HBr (b) After the addition of 15.9 mL of HBr (c) At the titration midpoint

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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How do I solve for this

**Determining pH During the Titration of Dimethylamine by HBr**

In this exercise, you are asked to determine the pH at various points during the titration of 39.7 mL of 0.222 M dimethylamine, \((\text{CH}_3)_2\text{NH}\), with a base ionization constant \(K_b\) of \(5.9 \times 10^{-4}\), by 0.222 M hydrobromic acid (HBr). The titration is conducted at 25 °C. State symbols are not shown for species in this problem.

Calculate the pH at the following stages:

(a) Before the addition of any HBr: [__________]

(b) After the addition of 15.9 mL of HBr: [__________]

(c) At the titration midpoint: [__________]

(d) At the equivalence point: [__________]

(e) After adding 61.5 mL of HBr: [__________]

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