Question

calculate the ph when 20.5 ml of 0.30 m hcl is titrated with 20.0 ml of 0.30 m naoh. round your answer to two decimal places.

Answer 1

To calculate the pH when 20.5 mL of 0.30 M HCl is titrated with 20.0 mL of 0.30 M NaOH, we need to determine the number of moles of HCl and NaOH and calculate the resulting concentration of the products. The pH of the resulting solution is 7.00.

Step-by-step explanation:

To calculate the pH when 20.5 mL of 0.30 M HCl is titrated with 20.0 mL of 0.30 M NaOH, we need to determine the number of moles of HCl and NaOH, and then use the balanced equation to find the resulting concentration of the products. Since HCl and NaOH react in a 1:1 ratio, we can use the following equation:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

First, we calculate the number of moles of HCl:

moles of HCl = volume of HCl x molarity of HCl = 0.0205 L x 0.30 mol/L = 0.00615 mol

Since HCl and NaOH react in a 1:1 ratio, the moles of NaOH will be the same as the moles of HCl.

Next, we calculate the resulting concentration of the products:

Concentration of NaCl = moles of NaCl / total volume of solution = 0.00615 mol / (0.0205 L + 0.0200 L) = 0.146 M

Finally, we can calculate the pH using the concentration of NaCl. Since NaCl is a neutral salt, it will not affect the pH significantly. Therefore, the pH will be determined by the concentration of H+ ions, which comes from the dissociation of water:

pH = -log[H+]

Since water dissociates into equal concentrations of H+ and OH-:

[H+] = [OH-] = 10-7 M

Therefore, the pH is:

pH = -log(10-7) = 7.00

Answer 2

pH= 2.43

Step-by-step explanation:

First, calculate the moles of acid in the solution:

(0.0205 L )(0.30molL)=0.00615 mol acid

Next, calculate the moles of base:

(0.0200 L)(0.30molL)=0.0060 mol base

Since the amount of acid has exceeded the amount of base, we can calculate the post-equivalence concentration of H3O+.

[H3O+]=nH3O+V=(0.00615−0.0060) mol H3O+(0.0205+0.0200) L[H3O+]≈0.003704 M

pH=−log(0.003704)=2.43

Since the hydronium concentration is precise to two significant figures, the logarithm should be rounded to two decimal places.