Question

When 14.7 mL of aqueous HBr was added to water, 482 mL of a solution with a pH of 4.23 was produced. What was the molarity of the initial HBr solution

Answer 1

Answer:

Molarity of initial HBr solution is $1.93\times 10^{-3}M$

Explanation:

$pH=-log[H^{+}]$  , where $[H^{+}]$ represents concentration of $H^{+}$ in molarity.

Here $pH=4.23$

So, $[H^{+}]=10^{-4.23}M=5.89\times 10^{-5}M$

Let's assume initial molarity of HBr is C (M) then initial concentration of $H^{+}$ will also be equal to C (M) as 1 mol of HBr produces 1 mol of $H^{+}$

So, in accordance with Laws of dilution ($C_{1}V_{1}=C_{2}V_{2}$):

$(14.7mL)\times [C(M)]=(482mL)\times [5.89\times 10^{-5}M]$

or, $C=1.93\times 10^{-3}$

Hence molarity of initial HBr solution is $1.93\times 10^{-3}M$