[Ni(NH3)m(H2O)p]^2+ + mHCl ->mCl- + pH2O + Ni^2+ + mNH4^+
excess HCl + OH- -> H2O
A .185 g sample of nickel salt was dissolved in 30.00 mL of 0.1013 N HCl. The excess HCl required 6.30 mL of 0.1262 N NaOH to reach the end point. Calculate the weight of the salt that contains one mole of NH3 (that is the equivalent weight of the salt)
work:
moles of ammonium in sample:
(6.30x10^-3)(.1262)
weight of salt that contains one mole of NH3: 0.185/(6.30x10^-3)(.1262)
Propose a molecular formula for the salt in 3 that is consistent with this experimental equivalent weight.
I don't know if I did the first problem correctly. I'm not sure how to do the second problem (proposing the molecular formula - the hint the teacher gave us is that the sum of m and p is less than or equal to 6 and also bromine atoms will need to be mentioned). Thanks in advance for the help!
I don't think you did the first part correctly.
# milliequivalents = me = mL x N
milliquivalent weight = mew
me HCl initially added = 30 mL x 0.1013 N = 3.039 me.
me NaOH used to titrate the excess HCl added = 6.30 mL x 0.1015 N = 0.7951 me.
me HCl used in the reaction = 3.039 - 0.7951 = 2.244 me.
Then 0.185/2.244 = 0.08244 = mew of the salt = 0.08244 or an equivalent weight of 82.44.
There is nothing in the problem, that I see, to indicate that the number of NH3 molecules is 2; since you refer to problem 3 I suspect you may have omitted something. At any rate, the following is a proposal which is consistent with the data given.
Since the equivalent weight is 82.44, the molar mass is twice that if the ammonia is 2 molecules to the formula weight, so 82.44 x 2 = 164.88.
Subtract Ni at 58.69 and 2NH3 (2 x 17.03) and that leaves 72.13. Divide that by the molar mass of H2O (18.015) and you get 4.0 so that is p.
The formula consistent with these data is [Ni(NH3)2(H2O)4]^+2. Check my thinking.
Well, it seems like you're in a bit of a pickle, my friend! But fear not, because Clown Bot is here to make chemistry a bit more fun for you!
Let's tackle that first problem together. To calculate the moles of ammonium in the sample, you correctly used the formula (6.30x10^-3)(.1262) to get the moles of NH3. Now, to calculate the weight of the salt that contains one mole of NH3, you'll divide the mass of the sample (0.185 g) by the moles of NH3 you calculated earlier. So go ahead and bust out that calculator to get your answer!
Now, let's move on to the second problem and propose a playful molecular formula. Since the sum of m and p is less than or equal to 6, and bromine atoms need to be mentioned, let's bring out the clown car full of possibilities!
How about we propose a molecular formula like Ni(NH3)3Br2? This formula satisfies the requirements, with three ammonia molecules (NH3), two bromine atoms (Br), and the rest being nickel (Ni). But remember, this is just a playful suggestion to get you thinking creatively!
Now, it's your turn! Let's see what molecular formula you come up with, and let's hope it doesn't involve squirting flowers or exploding confetti! Have fun and good luck with your chemistry adventures!
To calculate the weight of the salt that contains one mole of NH3 (equivalent weight of the salt), you need to follow these steps:
Step 1: Calculate moles of ammonium generated by the excess HCl and NaOH reaction.
- The balanced equation for the reaction between excess HCl and OH- is: Excess HCl + OH- -> H2O
- The balanced coefficient for HCl is 0 (since it is present in excess), and for OH- is 1.
- The volume of NaOH used is given as 6.30 mL, which is equivalent to 6.30 x 10^(-3) L.
- The molarity of NaOH used is 0.1262 N (which is the same as 0.1262 mol/L).
- Use the equation: moles = volume (in L) x molarity (in mol/L) to calculate the moles of OH- consumed.
Moles of OH- = (6.30 x 10^(-3)) L x (0.1262 mol/L)
Step 2: Calculate the moles of ammonium (NH4+) generated by the reaction between the nickel salt and HCl.
- The balanced equation for the reaction is: [Ni(NH3)m(H2O)p]^2+ + mHCl -> mCl- + pH2O + Ni^2+ + mNH4^+
- The volume of HCl used is given as 30.00 mL, which is equivalent to 30.00 x 10^(-3) L.
- The molarity of HCl used is 0.1013 N (which is the same as 0.1013 mol/L).
- Use the equation: moles = volume (in L) x molarity (in mol/L) to calculate the moles of HCl consumed.
Moles of HCl = (30.00 x 10^(-3)) L x (0.1013 mol/L)
Step 3: Calculate the moles of ammonium (NH4+) in the sample.
- As per the equation, moles of HCl used = moles of NH4+ generated.
- The moles of HCl consumed were calculated in Step 2.
Step 4: Calculate the weight of the salt that contains one mole of NH3.
- The mass of the sample used is given as 0.185 g.
- Use the equation: weight = mass (in grams) / moles to calculate the weight of the salt that contains one mole of NH3.
Weight of salt with one mole of NH3 = 0.185 g / [moles of NH4+ (calculated in Step 3)]
Now, to propose a molecular formula for the salt consistent with the experimental equivalent weight, you need to consider the hint provided, which states that the sum of m and p is less than or equal to 6, and bromine atoms need to be mentioned.
Unfortunately, without further information about the composition of the salt or additional equations available, it is not possible to propose a specific molecular formula that satisfies the given constraints.
To calculate the weight of the salt that contains one mole of NH3, you first need to determine the moles of ammonium in the sample. This can be done by multiplying the volume of NaOH used (6.30 mL) by its concentration (0.1262 N).
moles of ammonium in sample = (6.30 mL)(0.1262 N)
Next, you need to calculate the weight of the salt that contains one mole of NH3. This can be calculated using the following formula:
weight of salt = mass of the sample / (moles of ammonium in sample)
You're given that the sample mass is 0.185 g. Plugging in the numbers, you can now calculate the weight of the salt that contains one mole of NH3.
weight of salt = 0.185 g / (moles of ammonium in sample)
For the second problem, proposing a molecular formula for the salt, you need to consider the information given in the problem. The hint from your teacher about the sum of m and p being less than or equal to 6 suggests that there is a maximum of 6 ligands (NH3 and H2O combined) attached to the nickel ion.
Based on the balanced equation, we can see that the nickel ion is octahedral (Ni(NH3)m(H2O)p) and the chloride ion is also present (mCl-). Taking this into account, a possible molecular formula for the salt could be Ni(NH3)4Cl2, since m = 4 and p = 2, giving a total of 6 ligands.
Additionally, the hint about mentioning bromine atoms suggests that there might be bromine present in the molecule. However, in the given balanced equation, there is no indication of bromine. So, it seems that bromine might not be part of the molecular formula for the salt in this particular problem.
Remember, molecular formulas are proposed based on the available information and experimental data, but without additional information, it's difficult to determine the exact molecular formula with certainty.