Question

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sday - sem 1 - toronto dsb

question 18 (1 point)
solve for ( k_b ) if ( k_a = 1 cdot 10^{-3} )
( 10^3 )
( 1 cdot 10^{-7} )
( 10^{-11} )
3

question 19 (1 point)
borax and boric acid can be used to create a buffer at ph = 9.02.
the expression for this system is:
boric acid + water (
ightleftharpoons ) borax + ( h_3o^+ )
if 0.5 ml of 0.1 mol/l hydrochloric acid is added to this system, which reaction is favoured and how will the ph change?
reverse reaction, ph < 9.02
reverse reaction, ph = 9.02
forward reaction, ph = 9.02
forward reaction, ph < 9.02

question 20 (1 point)
polyprotic acids have multiple ( k_a ) values. what statement best describes this?

Explanation:

Question 18

Step1: Recall the relationship between \( K_a \), \( K_b \), and \( K_w \)

The relationship is \( K_a \times K_b = K_w \), where \( K_w = 1\times10^{-14} \) at 25°C.

Step2: Solve for \( K_b \)

We know \( K_a = 1\times10^{-3} \), so rearrange the formula to \( K_b=\frac{K_w}{K_a} \). Substitute the values: \( K_b = \frac{1\times10^{-14}}{1\times10^{-3}} \). Using the rule of exponents \( \frac{10^m}{10^n}=10^{m - n} \), we get \( K_b = 1\times10^{-11} \).

When HCl (a strong acid) is added, it provides \( H_3O^+ \) ions. The reaction given is boric acid + water \(
ightleftharpoons \) borax + \( H_3O^+ \). Adding \( H_3O^+ \) (from HCl) will increase the concentration of the product, so by Le Chatelier's principle, the reverse reaction will be favored (to consume the excess \( H_3O^+ \)). However, since it's a buffer, the pH change is minimal, but adding an acid (HCl) will make the solution slightly more acidic, so \( pH < 9.02 \).

Answer:

\( 10^{-11} \) (corresponding to the option with \( 10^{-11} \))

Question 19