Calculate the volume of 2.45 M NaI that would be needed to precipitate all of the Hg+2 ion from 185 mL of a 2.44 M Hg(NO3)2.
AND
Calculate the molarity of an acetic acid solution if 47.81 mL of the solution is needed to neutralize 147 mL of 1.32 M sodium hydroxide.
To calculate the volume of 2.45 M NaI needed to precipitate all of the Hg+2 ions, we need to first determine the stoichiometric ratio between NaI and Hg+2. The balanced equation for the reaction between NaI and Hg(NO3)2 is:
2 NaI + Hg(NO3)2 -> HgI2 + 2 NaNO3
From the equation, we can see that two moles of NaI react with one mole of Hg(NO3)2 to form one mole of HgI2.
Step 1: Calculate the number of moles of Hg(NO3)2 in the given solution:
Molarity of Hg(NO3)2 = 2.44 M
Volume of Hg(NO3)2 = 185 mL = 0.185 L
Number of moles of Hg(NO3)2 = Molarity x Volume
= 2.44 M x 0.185 L
Step 2: Determine the number of moles of NaI needed:
From the balanced equation, we see that the stoichiometric ratio between NaI and Hg(NO3)2 is 2:1. Therefore, the number of moles of NaI needed is twice the number of moles of Hg(NO3)2.
Number of moles of NaI = 2 x Number of moles of Hg(NO3)2
Step 3: Calculate the volume of 2.45 M NaI needed:
Molarity of NaI = 2.45 M
Volume of NaI = Number of moles of NaI / Molarity of NaI
Now, let's move on to the second question.
To calculate the molarity of the acetic acid solution, we need to use the balanced equation for the neutralization reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we see that the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1. Therefore, one mole of acetic acid reacts with one mole of sodium hydroxide.
Step 1: Calculate the number of moles of NaOH in the given solution:
Molarity of NaOH = 1.32 M
Volume of NaOH = 147 mL = 0.147 L
Number of moles of NaOH = Molarity x Volume
= 1.32 M x 0.147 L
Step 2: Determine the number of moles of acetic acid:
From the balanced equation, we see that the stoichiometric ratio between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is 1:1. Therefore, the number of moles of acetic acid is the same as the number of moles of sodium hydroxide.
Number of moles of acetic acid = Number of moles of NaOH
Step 3: Calculate the molarity of the acetic acid solution:
Volume of acetic acid solution = 47.81 mL = 0.04781 L
Molarity of acetic acid = Number of moles of acetic acid / Volume of acetic acid solution