1) You are given solutions of HCl and NaOH and must determine their concentrations. You use 37.0mL of NaOH to titrate 100mL of HCl and 13.6 mL of NaOH to titrate 50.0mL of 0.0782 M H2SO4. Find the unknown concentrations.
Molarity of NaOH and molarity of HCL?
2) Sodium hydroxide is used extensively in acid-base titrations because it is a strong, inexpensive base. A sodium hydroxide solution was standardized by titrating 28.00mL of 0.1598 M standard HCl. The initial buret reading of the sodium hydroxide was 2.54mL, and the final reading was 39.85mL. What is the molarity of the base solution?
#1. For H2SO4L:
2NaOH + H2SO4 --> Na2SO4 + 2H2O
moles H2SO4 = M x L = ??
moles NaOH = 2x moles H2SO4 (from the equation).
M NaOH = moles NaOH/L NaOH.
#1. For HCl:
NaOH + HCl ==> NaCl + H2O
moles NaOH = M(from the first part) x L
moles HCl = moles NaOH (from the equation).
M HCl = moles HCl/L HCl
#2.
NaOH + HCl ==> NaCl + H2O
moles HCl = M x L = ??
moles NaOH = moles HCl (from the equation).
M NaOH = moles NaOH/L NaOH
Note: volume of NaOH is determined by subtracting the initial reading from the final reading.
1) Well, it seems like you're dealing with some serious chemistry here! But don't worry, I'm here to add a little humor to your equations.
To find the unknown concentrations, let's break it down step by step.
First, let's focus on the titration of NaOH with HCl. You use 37.0 mL of NaOH to titrate 100 mL of HCl. Now, in this case, we're assuming a 1:1 stoichiometry between NaOH and HCl. So, based on that, we can set up the equation:
(moles of NaOH) / (volume of NaOH in liters) = (moles of HCl) / (volume of HCl in liters)
Let's plug in the values we know, and let "x" represent the molarity of NaOH:
(x mol/L) * (0.037 L) = (0.1 mol/L) * (1 L)
Solving for "x," we find that the molarity of NaOH is x = 0.37 M.
Now, let's move on to the second part where you titrate NaOH with H2SO4. You use 13.6 mL of NaOH to titrate 50.0 mL of 0.0782 M H2SO4. Following a similar approach as before, we can set up another equation:
(x mol/L) * (0.0136 L) = (0.0782 mol/L) * (0.05 L)
Solving for "x," we find that the molarity of NaOH is x = 0.2894 M.
Phew! Looks like you've got yourself some answers. The molarity of NaOH is 0.37 M in the first case and 0.2894 M in the second case.
1) To determine the unknown concentrations of NaOH and HCl, we can use the concept of stoichiometry and the balanced chemical equations for the neutralization reactions.
Let's start with the first titration:
HCl + NaOH -> NaCl + H2O
Using the volume of NaOH solution (37.0 mL), we can set up the following ratios:
(37.0 mL NaOH) / (100 mL HCl) = (x M NaOH) / (0.0782 M HCl)
x = (37.0 mL NaOH * 0.0782 M HCl) / (100 mL HCl)
x = 0.02891 M NaOH
Therefore, the molarity of NaOH is 0.02891 M.
Moving on to the second titration:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
Using the volume of NaOH solution (13.6 mL) and the molarity of H2SO4 (0.0782 M), we can set up the following ratios:
(13.6 mL NaOH) / (50.0 mL H2SO4) = (x M NaOH) / (0.0782 M H2SO4)
x = (13.6 mL NaOH * 0.0782 M H2SO4) / (50.0 mL H2SO4)
x = 0.0211 M NaOH
Therefore, the molarity of NaOH is 0.0211 M.
To summarize:
- The molarity of NaOH is 0.02891 M in the first titration.
- The molarity of NaOH is 0.0211 M in the second titration.
It is important to note that the molarity of HCl was not given in the problem. The molarity of HCl is not required to be determined for this question.
2) To find the molarity of the sodium hydroxide (NaOH) solution, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction:
HCl + NaOH -> NaCl + H2O
Using the initial and final buret readings for NaOH (2.54 mL and 39.85 mL, respectively), we can calculate the volume of NaOH used:
Volume of NaOH = Final reading - Initial reading
Volume of NaOH = 39.85 mL - 2.54 mL
Volume of NaOH = 37.31 mL
Using the volume of NaOH (37.31 mL) and the molarity of HCl (0.1598 M), we can set up the following ratio:
(37.31 mL NaOH) / (28.00 mL HCl) = (x M NaOH) / (0.1598 M HCl)
x = (37.31 mL NaOH * 0.1598 M HCl) / (28.00 mL HCl)
x = 0.21045 M NaOH
Therefore, the molarity of the base solution (NaOH) is 0.21045 M.
1) To determine the unknown concentrations, we can use the concept of stoichiometry and the equation for neutralization reactions.
Let's start with the first titration where 37.0 mL of NaOH was used to titrate 100 mL of HCl. We can set up the following balanced equation:
NaOH + HCl -> NaCl + H2O
From the balanced equation, we can see that the ratio between NaOH and HCl is 1:1. This means that 37.0 mL of NaOH is equal to 100 mL of HCl in terms of their stoichiometric amounts.
Now, let's move to the second titration where 13.6 mL of NaOH was used to titrate 50.0 mL of 0.0782 M H2SO4. Again, we can set up the balanced equation:
NaOH + H2SO4 -> Na2SO4 + H2O
From the balanced equation, we can see that the ratio between NaOH and H2SO4 is 2:1. This means that 13.6 mL of NaOH is equal to half the stoichiometric amount of H2SO4 (25 mL).
To find the unknown concentrations, we can use the stoichiometric ratios and the given information.
For HCl, 37.0 mL of NaOH is equivalent to 100 mL of HCl. Therefore, we can say that the concentration of HCl is the same as the concentration of NaOH used in this titration.
For H2SO4, 13.6 mL of NaOH is equivalent to 25 mL of H2SO4. We have the volume (50 mL) and concentration (0.0782 M) of H2SO4. We can set up the following equation using the concentration formula:
C1V1 = C2V2
(0.0782 M) x (25 mL) = C2 x (50 mL)
Solving for C2 (the concentration of NaOH), we get C2 = (0.0782 M) x (25 mL) / (50 mL) = 0.0391 M.
Therefore, the molarity of NaOH is 0.0391 M and the molarity of HCl is also 0.0391 M.
2) In this problem, we are given the initial buret reading of the sodium hydroxide (2.54 mL), the final reading (39.85 mL), and the volume of HCl used in the titration (28.00 mL).
To find the moles of HCl used, we can use the concentration and volume of HCl:
moles of HCl = concentration x volume = (0.1598 M) x (28.00 mL)
Next, we need to determine the mole ratio between NaOH and HCl. From the balanced equation:
NaOH + HCl -> NaCl + H2O
we can see that the ratio is 1:1. This means that the moles of NaOH used is equal to the moles of HCl used.
Now, let's find the moles of NaOH used:
moles of NaOH = moles of HCl
Finally, we can calculate the molarity of the base solution (NaOH).
Molarity of NaOH = moles of NaOH / volume of NaOH = moles of HCl / final volume of NaOH
Moles of HCl = (0.1598 M) x (28.00 mL)
Final volume of NaOH = 39.85 mL - 2.54 mL = 37.31 mL
Molarity of NaOH = (0.1598 M) x (28.00 mL) / (37.31 mL)
Solving for the molarity, we can obtain the desired result.