answer:To find the pH of a 0.1 M solution of acetic acid, we can use the formula for the dissociation constant (Ka) and the definition of pH. The dissociation of acetic acid can be represented as: CH3COOH ⇌ CH3COO- + H+ The Ka expression for this reaction is: Ka = [CH3COO-][H+]/[CH3COOH] Since the initial concentration of acetic acid is 0.1 M, we can represent the change in concentration as follows: CH3COOH: 0.1 - x CH3COO-: x H+: x Now, we can substitute these values into the Ka expression: 1.8 x 10^-5 = (x)(x)/(0.1 - x) Assuming that x is much smaller than 0.1, we can simplify the equation: 1.8 x 10^-5 ≈ x^2/0.1 Now, we can solve for x: x^2 = 1.8 x 10^-5 * 0.1 x^2 = 1.8 x 10^-6 x ≈ √(1.8 x 10^-6) x ≈ 1.34 x 10^-3 Since x represents the concentration of H+ ions, we can now find the pH using the definition of pH: pH = -log10[H+] pH = -log10(1.34 x 10^-3) pH ≈ 2.87 Therefore, the pH of the 0.1 M solution of acetic acid is approximately 2.87.

answer:In this nucleophilic substitution reaction, the nucleophile is the hydroxide ion (OH-) from sodium hydroxide (NaOH), and the electrophile is methyl bromide (CH3Br). The hydroxide ion will attack the electrophilic carbon in methyl bromide, replacing the bromide ion (Br-). The products of this reaction are methanol (CH3OH) and sodium bromide (NaBr). The balanced chemical equation for this reaction is: CH3Br + NaOH → CH3OH + NaBr

answer:When 1-bromobutane is reacted with sodium methoxide in methanol, a nucleophilic substitution reaction called SN2 (Substitution Nucleophilic Bimolecular) will occur. The mechanism of this reaction is as follows: 1. Sodium methoxide (CH3O-Na+) dissociates into a methoxide ion (CH3O-) and a sodium ion (Na+). The methoxide ion acts as a strong nucleophile. 2. The methoxide ion (CH3O-) attacks the electrophilic carbon atom (the one bonded to the bromine) in 1-bromobutane from the backside, opposite to the leaving group (Br-). 3. As the methoxide ion forms a bond with the electrophilic carbon, the carbon-bromine bond starts to break, and the bromide ion (Br-) is released as a leaving group. 4. The product of this reaction is 1-methoxybutane, and the reaction proceeds with inversion of configuration at the electrophilic carbon due to the backside attack of the nucleophile. The overall reaction can be written as: CH3CH2CH2CH2Br + CH3O-Na+ → CH3CH2CH2CH2OCH3 + NaBr

answer:The reaction between 1-bromopropane and sodium ethoxide in ethanol is an SN2 (nucleophilic substitution bimolecular) reaction. In this reaction, the ethoxide ion (CH3CH2O-) from sodium ethoxide acts as a nucleophile and attacks the electrophilic carbon (the one attached to the bromine) in 1-bromopropane. The bromide ion (Br-) leaves as a leaving group, and sodium bromide is formed as a byproduct. The major product of this reaction is propyl ethyl ether (CH3CH2CH2OCH2CH3). The reaction can be represented as follows: CH3CH2CH2Br + CH3CH2ONa → CH3CH2CH2OCH2CH3 + NaBr