Question

What volume of a 0.500 M HCl solution is needed to neutralize each: 14.0 ml of a 0.300 M NaOH and 19.0ml of a 0.200 M Ba(OH)2

Answer 1

To neutralize 14.0 ml of 0.300 M NaOH, 8.4 mL of 0.500 M HCl solution is needed.

To neutralize 19.0 ml of 0.200 M Ba(OH)2, 15.2 mL of 0.500 M HCl solution is needed.

Step-by-step explanation:

To find the volume of 0.500 M HCl solution needed to neutralize each solution, we use the balanced chemical equation:

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

For the neutralization of 14.0 mL of 0.300 M NaOH, we can use the equation to determine the moles of NaOH:

0.300 M NaOH × 0.0140 L NaOH = 0.0042 mol NaOH

Since the ratio of HCl to NaOH is 1:1, we know that 0.0042 mol of HCl is required.

Using the definition of molarity, we can calculate the volume of 0.500 M HCl needed:

Volume of 0.500 M HCl = 0.0042 mol HCl / 0.500 M HCl = 0.0084 L HCl = 8.4 mL HCl

Similarly, for the neutralization of 19.0 mL of 0.200 M Ba(OH)2, we can calculate the moles of Ba(OH)2:

0.200 M Ba(OH)2 × 0.0190 L Ba(OH)2 = 0.0038 mol Ba(OH)2

Using the 1:2 ratio of HCl to Ba(OH)2, we know that 2 mol of HCl are required for every 1 mol of Ba(OH)2.

Thus, 2 × 0.0038 mol = 0.0076 mol of HCl is required.

Using the definition of molarity, we can calculate the volume of 0.500 M HCl needed:

Volume of 0.500 M HCl = 0.0076 mol HCl / 0.500 M HCl = 0.0152 L HCl = 15.2 mL HCl

Answer 2

First thing, we need to know the chemical equation for each neutralization reaction. Then, determine the amount in moles needed hydrochloric acid. We do as follows:

HCl + NaOH = NaCl + H2O

0.300 M NaOH ( .014 L ) ( 1 mol HCl / 1 mol NaOH ) ( 1 L / 0.500 M HCl ) = 0.0084 L HCl solution or 8.4 mL of 0.500 M HCl

2HCl + Ba(OH)2 = BaCl2 + 2H2O

0.200 M Ba(OH)2 (0.019 L ) ( 2 mol HCl / 1 mol Ba(OH)2 ) ( 1 L / 0.500 M HCl ) = 0.0152 L HCl solution or 15.2 mL of 0.500 M HCl