Question

What is the molarity of the solution of Sodium hydroxide, if 35 ml of 0.1 M HCl is used in the titration of 25 ml of the barium hydroxide solution

Answer 1

To determine the molarity of the sodium hydroxide solution, calculate the moles of HCl that reacted, which also equals the moles of NaOH due to the 1:1 stoichiometry, and then divide by the volume of NaOH used. The molarity of NaOH is 0.14 M.

Step-by-step explanation:

To find the molarity of the sodium hydroxide (NaOH) solution used to titrate 35 mL of 0.1 M hydrochloric acid (HCl), we need to use the concept of molarity and the stoichiometry of the balanced chemical equation between HCl and NaOH, which is:

HCl + NaOH → NaCl + H₂O

First, calculate the number of moles of HCl:

Moles of HCl = volume (L) × molarity (M)

Moles of HCl = 0.035 L × 0.1 M = 0.0035 moles

Since the molar ratio between HCl and NaOH is 1:1, the moles of NaOH will also be 0.0035. Next, we calculate the molarity of NaOH using the volume of NaOH:

Molarity of NaOH = moles of NaOH / volume of NaOH (L)

Molarity of NaOH = 0.0035 moles / 0.025 L = 0.14 M

The molarity of the sodium hydroxide solution is therefore 0.14 M.

Answer 2

Answer: The molarity of
`Ba(OH)_2` is 0.07 M

Step-by-step explanation:

According to the neutralization law,

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1` = basicity
`HCl` = 1

`M_1` = molarity of
`HCl` solution = 0.1 M

`V_1` = volume of
`HCl` solution = 35 ml

`n_2` = acidity of
`Ba(OH)_2` = 2

`M_2` = molarity of
`Ba(OH)_2` solution = ?

`V_2` = volume of
`Ba(OH)_2` solution = 25 ml

Putting in the values we get:

`1* 0.1* 35=2* M_2* 25`

`M_2=0.07`

Therefore, molarity of
`Ba(OH)_2` is 0.07 M