Question

the ka of an acid is ..... to the kb of its conjugate base. equal inversely proportional directly proportional unrelated question 5 what is the value of pk_w at 25°c? 10 14 7 0 question 6 for a conjugate acid - base pair, if the k_a is known, how can k_b be calculated? by adding k_a to k_w by dividing k_w by k_a by multiplying k_a with k_w by subtracting k_a from k_w

Explanation:

Question 1

• For a conjugate acid - base pair, the relationship between \(K_a\) (acid dissociation constant) and \(K_b\) (base dissociation constant) is given by \(K_a\times K_b = K_w\). If we rewrite this as \(K_b=\frac{K_w}{K_a}\), we can see that as \(K_a\) increases, \(K_b\) decreases and vice - versa. So, the \(K_a\) of an acid is inversely proportional to the \(K_b\) of its conjugate base.

Question 2

• The ion - product constant for water, \(K_w=[H^+][OH^-]\). At \(25^{\circ}C\), \(K_w = 1.0\times10^{-14}\). The formula for \(pK_w\) is \(pK_w=-\log_{10}(K_w)\). Substituting \(K_w = 1.0\times 10^{-14}\) into the formula, we get \(pK_w=-\log_{10}(1.0\times10^{-14})=14\).

Question 3

• We know from the relationship \(K_a\times K_b=K_w\). To solve for \(K_b\), we can rearrange the formula as \(K_b=\frac{K_w}{K_a}\).

Answer:

• Question 1: inversely proportional
• Question 2: 14
• Question 3: By dividing \(K_W\) by \(K_a\)