a) A standard solution of 0.243 M NaOH was used to determine the concentration of a hydrochloric acid solution. If 46.33 mL of NaOH is needed to neutralize 10.00 mL of the acid, what is the molar concentration of the acid?
b) A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration of salts as normal human blood. Calculate the mass of solute needed to prepare 275.0 mL of a physiological saline solution.
a) To find the molar concentration of the acid solution, we can use the equation:
Moles of NaOH = Moles of acid
First, calculate the moles of NaOH used:
Moles of NaOH = (0.243 M) x (46.33 mL) / 1000 mL/L
= 0.0113 moles
Since NaOH and HCl react in a 1:1 ratio, the moles of HCl present in the acid solution are also 0.0113 moles.
Now, calculate the molar concentration of the acid solution:
Molar concentration of acid = Moles of acid / Volume of acid solution in liters
Volume of acid solution = 10.00 mL = 10.00 mL / 1000 mL/L = 0.01000 L
Molar concentration of acid = 0.0113 moles / 0.01000 L
= 1.13 M
b) To calculate the mass of solute needed for the physiological saline solution, we can use the equation:
Mass = Molarity x Volume x Molar mass
First, determine the moles of sodium chloride (NaCl) required:
Moles of NaCl = Molarity x Volume in liters
Molarity = 0.150 M
Volume = 275.0 mL = 275.0 mL / 1000 mL/L = 0.2750 L
Moles of NaCl = (0.150 M) x (0.2750 L)
= 0.0413 moles
The molar mass of sodium chloride (NaCl) is 58.44 g/mol.
Now, calculate the mass of NaCl needed:
Mass of NaCl = Moles of NaCl x Molar mass
Mass of NaCl = (0.0413 moles) x (58.44 g/mol)
= 2.413 g
Therefore, you would need 2.413 grams of sodium chloride to prepare 275.0 mL of a physiological saline solution.
a) To determine the molar concentration of the hydrochloric acid solution, we can use the concept of stoichiometry and the equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
First, let's find the number of moles of NaOH used during the neutralization:
Number of moles of NaOH = volume of NaOH solution (in L) * molarity of NaOH
Since the volume of NaOH solution is given as 46.33 mL, we need to convert it to liters:
Volume of NaOH solution = 46.33 mL * (1 L / 1000 mL) = 0.04633 L
Now we can calculate the number of moles:
Number of moles of NaOH = 0.04633 L * 0.243 mol/L = 0.01128399 mol
According to the balanced equation, 1 mole of NaOH reacts with 1 mole of HCl. Therefore, the number of moles of HCl is also 0.01128399 mol.
Finally, we can determine the molar concentration of the hydrochloric acid solution:
Molar concentration of HCl = number of moles of HCl / volume of HCl solution (in L)
The volume of the acid solution is given as 10.00 mL, which we convert to liters:
Volume of HCl solution = 10.00 mL * (1 L / 1000 mL) = 0.010 L
Now we can calculate the molar concentration:
Molar concentration of HCl = 0.01128399 mol / 0.010 L = 1.128399 M
Therefore, the molar concentration of the hydrochloric acid solution is approximately 1.128 M.
b) To calculate the mass of solute needed to prepare a physiological saline solution, we need to know the molar mass of sodium chloride (NaCl) and the desired molar concentration of the solution.
The molar mass of NaCl is approximately 58.44 g/mol.
First, we need to calculate the number of moles of NaCl required:
Number of moles = molar concentration * volume of solution (in L)
The volume of solution is given as 275.0 mL, which we convert to liters:
Volume of solution = 275.0 mL * (1 L / 1000 mL) = 0.275 L
Now we can calculate the number of moles:
Number of moles = 0.150 mol/L * 0.275 L = 0.04125 mol
Finally, we can determine the mass of NaCl needed using the molar mass:
Mass of NaCl = number of moles * molar mass
Mass of NaCl = 0.04125 mol * 58.44 g/mol = 2.4075 g
Therefore, to prepare 275.0 mL of a physiological saline solution with a molar concentration of 0.150 M, you would need approximately 2.4075 grams of sodium chloride.
HCl + NaOH ==> NaCl + H2O
mols NaOH = M x L = ?
From the equation you can see mols HCl \= mols NaOH
Then M HCl = mols HCl/L HCl
2.
0.150M = 0.150 mols/L = (0.150*molar mass NaCl)/L = about 8.9 g NaCl/L but that's an approx and you should redo this problem more accurately.
But you don't need 1L, you need only 275.0 mL; therefore,
8.9 g NaCl x (275/1000) = ? g NaCl needed for 275 mOl.
The easier way to work it, I think, is to ask yourself how many mols you need in the new solution? That's M x L = 0.15 x 0.275 = ?
Then ? x molar mass NaCl = g NaCl.