. A student added 40.0 mL of an NaOH solution to 90.0 mL of 0.400 M HCl. The
solution was then treated with an excess of nickel(II) nitrate, resulting in the formation
of 1.06 g of Ni(OH)2 precipitate. Determine the concentration of the original NaOH
solution.
Convert 1.06 g Ni(OH)2 to g NaOH. That's
1.06 x (2*molar mass NaOH/molar mass Ni(OH)2) = ?. That's the g NaOH in excess.
How many g NaOH was in the original? That must be g excess NaOH + g titrated by HCl. How much was titrated by HCl? That's L x M x molar mass NaOH = 0.090 x 0.40 x 40 = ? total g NaOH
Convert to mols NaOH, then M = mols/L.
Post your work if you get stuck.
To determine the concentration of the original NaOH solution, we can use the concept of stoichiometry and the amount of Ni(OH)2 precipitate formed. Here's the step-by-step solution:
Step 1: Write the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
Step 2: Determine the moles of Ni(OH)2 precipitate formed:
Given mass of Ni(OH)2 = 1.06 g
Molar mass of Ni(OH)2 = 62.69 g/mol
Moles of Ni(OH)2 = (1.06 g) / (62.69 g/mol)
Step 3: Determine the moles of NaOH used in the reaction:
From the balanced equation, the stoichiometric ratio between NaOH and Ni(OH)2 is 2:1.
Therefore, moles of NaOH = 2 * moles of Ni(OH)2
Step 4: Determine the volume of the NaOH solution used:
Given volume of NaOH solution = 40.0 mL = 0.040 L
Step 5: Calculate the concentration of NaOH in the original solution:
Concentration (C) = moles / volume
Using the given values in Step 2 and Step 4, we can calculate the concentration of NaOH as follows:
C = (2 * moles of Ni(OH)2) / volume
Plug in the numbers:
C = (2 * (1.06 g / 62.69 g/mol)) / 0.040 L
Simplify:
C = (2 * 0.0169 mol) / 0.040 L
C = 0.845 M
Therefore, the concentration of the original NaOH solution is 0.845 M.
To determine the concentration of the original NaOH solution, we can use the concept of stoichiometry and titration.
Here's the step-by-step process to solve the problem:
1. First, let's write the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
2. Next, calculate the number of moles of HCl used in the reaction. Since the volume is given in milliliters (mL), we need to convert it to liters (L):
Volume of HCl = 90.0 mL = 0.090 L
Concentration of HCl = 0.400 M (moles per liter)
Number of moles of HCl = Concentration × Volume = 0.400 M × 0.090 L
3. Since the balanced equation has a 1:1 ratio between HCl and NaOH, the number of moles of NaOH used in the reaction is also equal to the moles of HCl.
4. Now, calculate the number of moles of NaOH used:
Number of moles of NaOH = Number of moles of HCl = 0.400 M × 0.090 L
5. The next step is to calculate the concentration of NaOH. We know that 1.06 g of Ni(OH)2 precipitate was formed. From the balanced equation, we can see that the ratio between Ni(OH)2 and NaOH is 1:2. So, we need to calculate the number of moles of Ni(OH)2 and then convert it to moles of NaOH:
Molar mass of Ni(OH)2 = atomic mass of Ni + (2 × atomic mass of O) + (2 × atomic mass of H)
Number of moles of Ni(OH)2 = Given mass (1.06 g) / Molar mass of Ni(OH)2
Number of moles of NaOH = 2 × Number of moles of Ni(OH)2 (from the balanced equation)
6. Finally, we can calculate the concentration of NaOH:
Volume of NaOH = 40.0 mL = 0.040 L
Concentration of NaOH = Number of moles of NaOH / Volume of NaOH
By following these steps, you should be able to determine the concentration of the original NaOH solution.