Could someone please help me with these two questions. I am having trouble.
1. What is the molarity of a solution if 48.0 mL of a 0.220 M solution is required to neutralize a 25.0-mL sample of the solution?
2. The mixing of which pair of reactants will result in a precipitation reaction?
K 2SO4(aq) + Hg2 (NO3)2(aq)
CsI(aq) + NaOH(aq)
HCl(aq) + Ca(OH)2(aq)
NaNO3(aq) + NH4Cl(aq)
1. Hmmmm. Cannot be determined. Assuming one is acid and one is base, the immediate question is that a mole does not neutralize a mole: one could be di or tri protic. You have to have a balanced equation. Frankly, I like to work with Normal solution on these.
2. Look at your solubility rules, for instance: on the first, all potassium compounds are soluble, but not all mercury: is Mercury(II)sulfate soluble?
Memorize those rules, millions before you have done the same.
Ok, great! Thank you.
Sorry about question 1, I left out the full question. This is it:
What is the molarity of a NaOH solution if 48.0 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0-mL sample of NaOH solution?
2NaOH + H2SO4 --> Na2SO4 + H2O
moles H2SO4 = M x L
moles NaOH = 2x moles H2SO4
Molarity NaOH = moles NaOH/L NaOH.
Sure, I can help you with these two questions. Let's go step by step.
1. To find the molarity of a solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
In this case, we are given the volume of the solution (48.0 mL) and the molarity of the solution (0.220 M). However, we need to find the moles of solute.
The number of moles can be calculated using the formula:
Moles = Molarity x Volume (in liters)
First, we need to convert the volume of the solution from milliliters to liters:
48.0 mL = 48.0 / 1000 = 0.048 L
Now, we can find the moles of solute using the formula:
Moles = 0.220 M x 0.048 L = 0.01056 moles
Finally, we can calculate the molarity of the solution using the formula:
Molarity = 0.01056 moles / 0.025 L (given volume of sample) = 0.4224 M
Therefore, the molarity of the solution is 0.4224 M.
2. To determine which pair of reactants will result in a precipitation reaction, we need to identify if a precipitate will form when the two solutions are mixed. This can be done by looking at the solubility rules.
According to the solubility rules, generally, nitrates (NO3-) and alkali metal ions (e.g. Na+, K+, Cs+) are soluble, so NaNO3(aq) will not form a precipitate.
HCl(aq) and Ca(OH)2(aq) react to form H2O and CaCl2, which is a soluble compound, so no precipitate will form.
K2SO4(aq) and Hg2(NO3)2(aq) react to form 2KNO3 and Hg2SO4. According to the solubility rules, sulfates (SO4^2-) are generally soluble except when paired with mercury (Hg2^2+), which forms a slightly soluble compound. Therefore, a precipitate will form in this reaction.
CsI(aq) and NaOH(aq) will not form a precipitate because both compounds are soluble.
Therefore, the mixing of K2SO4(aq) and Hg2(NO3)2(aq) will result in a precipitation reaction.
I hope this helps! Let me know if you have any further questions.