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(a) Define (i) an acid, (ii) a base according to the Brønsted-Lowry theory - Leaving Cert Chemistry - Question 8 - 2003

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(a) Define (i) an acid, (ii) a base according to the Brønsted-Lowry theory. Identify the acid, and conjugate acid in the following system. $$ H_2S + OH^- \rightlef... show full transcript

Worked Solution & Example Answer:(a) Define (i) an acid, (ii) a base according to the Brønsted-Lowry theory - Leaving Cert Chemistry - Question 8 - 2003

Step 1

Define (i) an acid

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Answer

According to the Brønsted-Lowry theory, an acid is defined as a proton donor. This means that an acid is a substance that can donate a hydrogen ion (H⁺) to another substance in a chemical reaction.

Step 2

Define (ii) a base

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Answer

In the context of the Brønsted-Lowry theory, a base is defined as a proton acceptor. This means that a base is a substance that can accept a hydrogen ion (H⁺) from an acid during a chemical reaction.

Step 3

Identify the acid, and conjugate acid in the following system

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Answer

In the given equilibrium:

H2S+OHHS+H2OH_2S + OH^- \rightleftharpoons HS^- + H_2O The acid is H2SH_2S (as it donates a proton), and the conjugate acid is HSHS^- (the species formed after the acid donates a proton).

Step 4

Define pH

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Answer

pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration

pH=extlog10[H+]pH = - ext{log}_{10}[H^+] This measures the acidity or basicity of a solution.

Step 5

Calculate the approximate pH of the vinegar solution

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Answer

Given that the concentration of acetic acid is 6% (w/v), we can convert this to molarity.

  1. Calculation of molarity:

    • 6 g of acetic acid in 100 mL solution means:
    • Molar mass of acetic acid (C₂H₄O₂) = 60 g/mol.
    • Molarity (M) = rac{6 ext{ g}}{60 ext{ g/mol}} imes rac{1000 ext{ mL}}{100 ext{ mL}} = 1 ext{ mol/L}
  2. The dissociation constant, Ka=1.8imes105K_a = 1.8 imes 10^{-5}, can be used to find the concentration of hydrogen ions. We use the approximation:

    • Ka=[H+][A]/[HA]K_a = [H^+][A^-]/[HA] Assuming [H+]=[A][H^+] = [A^-] and [HA]extisapproximately1extmol/L[HA] ext{ is approximately } 1 ext{ mol/L}, we get:
    • 1.8imes105=[H+]2/(1)1.8 imes 10^{-5} = [H^+]^2 / (1)
    • [H+]=1.8imes105extwhichgivesapproximately0.00424extmol/L.[H^+] = \sqrt{1.8 imes 10^{-5}} ext{ which gives approximately } 0.00424 ext{ mol/L}.
  3. Finally, we can calculate pH:

    • pH=extlog10(0.00424)extwhichresultsinapproximately2.37.pH = - ext{log}_{10}(0.00424) ext{ which results in approximately } 2.37.

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