Question

A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.

Answer 1

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

$HBr+KOH\rightarrow KBr+H_2O$

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

$pOH=-log([OH^-])=-log(0.320)=0.50$

Thus, the pH is:

$pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50$

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

$n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol$

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

$n_{KOH}^{remaining}=0.00425mol$

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

$[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M$

So the pOH and the pH turn out:

$pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15$

Best regards!