Problems #1 - 10

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I answered this Yahoo Answers question:

"When H_{2}SO_{4}and NaOH is reacted, is it the H^{+}and OH¯ that has equal number of moles, or the H_{2}SO_{4}and NaOH has equal number of moles?"

I think it's a good question and I wanted to show you the question first, if you wanted to ponder it before looking at my answer.

**Problem #1:** A volume of 20.00 mL of 0.250 M Al(OH)_{3} neutralizes a 75.00 mL sample of H_{2}SO_{4} solution. What is the concentration of H_{2}SO_{4}?

**Solution #1:**

2Al(OH)_{3}+ 3H_{2}SO_{4}---> Al_{2}(SO_{4})_{3}+ 6H_{2}OAl(OH)

_{3}and H_{2}SO_{4}are in a 2 : 3 molar ratio.moles Al(OH)

_{3}---> (0.250 mol/L) (20.00 mL) = 5.00 millimoles2 is to 3 as 5.00 is to x

x = 7.50 millimoles

molarity H

_{2}SO_{4}---> 7.50 millimoles / 75.00 mL = 0.100 M

**Solution #2:**

(x) (75.0 mL) (0.250 M) (20.0 mL) ––––––––––– = ––––––––––––––––– 3 2 (3) (0.250 M) (20.0 mL) = (2) (x) (75.0 mL)

x = 0.100 M

**Problem #2:** A 21.62 mL sample of Ca(OH)_{2} solution was titrated with 0.2545 M HCl. 45.87 mL of the acid was required to reach the endpoint of the titration. What was the molarity of calcium hydroxide solution?

**Solution:**

2HCl + Ca(OH)_{2}---> CaCl_{2}+ 2H_{2}Omoles HCl ---> (0.2545 mol/L) (0.04587 L) = 0.011674 mol

use the 2 : 1 molar ratio of HCl to Ca(OH)

_{2}:2 is to 1 as 0.011674 mol is to xx = 0.005837 mol of Ca(OH)

_{2}molarity of Ca(OH)

_{2}---> 0.005837 mol / 0.02162 L = 0.2700 M (to four sig figs)

**Problem #3:** Calculate the volume of NaOH necessary to neutralize 50.0 mL of a 16.0 M solution of sulfuric acid. The concentration of the NaOH is 2.50 M.

**Solution:**

2NaOH + H_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}OCalculate moles of H

_{2}SO_{4}by using n = C x V:n = 16.0 mol/L x 50 mL = 800 millimolesNow look at the equation. It says that for every mole of H

_{2}SO_{4}you need twice that many moles of NaOH to neutralize all the H_{2}SO_{4}.So moles of NaOH = 800 x 2 = 1600 millimoles of NaOH needed.

Now that you have moles and a concentration you can find volume:

V = moles / concentrationV = 1600 millimoles / 2.50 mmoles/mL = 640. mL

Note: I copied this answer from Yahoo Answers and the original writer put the molarity as 2.50 moles/L. If you did that on an answer to a test, you might get a deduction for using the wrong units on the molarity. Moles does not cancel with millimoles, although the numerical answer is correct.

**Problem #4:** What is the citric acid concentration in a soda if it requires 32.27 mL of 0.0148 M NaOH to titrate 25.00 mL of soda?

**Solution:**

Citric acid has three acidic hydrogens, so I will use H_{3}Cit for the formula.H

_{3}Cit + 3NaOH ---> Na_{3}Cit + 3H_{2}O

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

(x) (25.0 mL) (0.0148 M) (32.27 mL) ––––––––––– = ––––––––––––––––– 1 3 (3) (x) (25.0 mL) = (1) (0.0148 M) (32.27 mL)

x = 0.00637 M (to three sig figs)

**Problem #5:** Consider the unbalanced molecular equation in which citric acid, H_{3}C_{6}H_{5}O_{7}, is reacted with sodium hydroxide:

H_{3}C_{6}H_{5}O_{7}(aq) + NaOH(aq) ---> Na_{3}C_{6}H_{5}O_{7}(aq) + H_{2}O(ℓ)

If 62.7 mL of 1.20 M NaOH(aq) is titrated with 32.0 mL of H_{3}C_{6}H_{5}O_{7}(aq) what the molarity of H_{3}C_{6}H_{5}O_{7}(aq)?

**Solution:**

1) First, let us balance the equation:

H_{3}C_{6}H_{5}O_{7}(aq) + 3NaOH(aq) ---> Na_{3}C_{6}H_{5}O_{7}(aq) + 3H_{2}O(ℓ)

2) The key point is the 1:3 stoichiometric ratio between the citric acid and the NaOH.

moles NaOH ---> (1.20 mol/L) (0.0627 L) = 0.07524 mol3) Calculate the molarity of the citric acid:You need three moles of NaOH to neutralize every one mole of citric acid.

moles citric acid ---> 0.07524 mol / 3 = 0.02508 mol

0.02508 mol / 0.0320 L = 0.78375 MThree sig figs gives 0.784 M for the final answer.

**Problem #6:** Calculate the molar concentration of phosphoric acid if 90.0 mL are required to neutralize 200. mL of 0.200 M calcium hydroxide solution.

**Solution:**

1) First, let us balance the equation:

2H_{3}PO_{4}(aq) + 3Ca(OH)_{2}(aq) ---> Ca_{3}(PO_{4})_{2}(s) + 6H_{2}O(ℓ)Notice that solid calcium phosphate is formed. This is immaterial to the following calculation.

2) There is a 2:3 molar ratio between H_{3}PO_{4} and Ca(OH)_{2}.

moles Ca(OH)_{2}---> (0.2 mol/L) (0.2 L) = 0.04 mol2 is to 3 as x is to 0.04

x = 0.02667 mol of H

_{3}PO_{4}

3) Calculate the molarity of the H_{3}PO_{4}:

0.02667 mol / 0.09 L = 0.296296 MRounded off to three sig figs, 0.296 M is the H

_{3}PO_{4}concentration.

Tetrahydroxyarsoranyl is an actual compound (I think) and its formula can be rendered as H_{4}AsO_{4}. The ChemTeam does not know if those four hydrogens are acidic hydrogens, but he's going to act like they are.

**Problem #7:** 35.0 mL of a solution of H_{4}AsO_{4} is titrated with 0.0840 M Ba(OH)_{2}. 41.4 mL of the base is required to reach the endpoint. Calculate the molarity of the H_{4}AsO_{4} solution.

**Solution:**

H_{4}AsO_{4}+ 2Ba(OH)_{2}---> Ba_{2}AsO_{4}+ 4H_{2}O

(x) (35.0 mL) (0.0840 M) (41.4 mL) ––––––––––– = ––––––––––––––––– 1 2 (1) (0.0840 M) (41.4 mL) = (2) (x) (35.0 mL)

x = 0.0497 M

**Problem #8:** 35.0 mL of a solution of H_{4}AsO_{4} is titrated with 0.0840 M Al(OH)_{3}. 41.4 mL of the base is required to reach the endpoint. Calculate the molarity of the H_{4}AsO_{4} solution.

**Solution:**

3H_{4}AsO_{4}+ 4Al(OH)_{3}---> Al_{4}(AsO_{4})_{3}+ 12H_{2}O

(x) (35.0 mL) (0.0840 M) (41.4 mL) ––––––––––– = ––––––––––––––––– 3 4 (3) (0.0840 M) (41.4 mL) = (4) (x) (35.0 mL)

x = 0.0745 M

**Problem #9:** A flask containes 30.0 mL of 0.100 M hydrochloric and 20.0 mL of hydrobromic acid. If 50.0 mL of 0.200 M sodium hydroxide was required to neutralize the mixture of acids in the flask, what was the concentration of the hydrobromic acid?

**Solution:**

**Solution using steps:**

moles of NaOH ---> (0.200 mol/L) (0.0500 L) = 0.0100 molNaOH and HCl react in a 1:1 molar ratio. Let's see how much NaOH is left over after reacting with the HCl.

moles HCl ---> (0.100 mol/L) (0.0300 L) = 0.00300 mol

0.0100 mol minus 0.00300 mol = 0.00700 mol <--- left over NaOH

NaOH and HBr react in a 1:1 molar ratio. To achieve neutralization, 0.00700 mol of NaOH reacts with 0.00700 mole of HBr.

molarity ---> 0.00700 mol / 0.0200 L = 0.350 M

**Solution using one equation:**

M_{a1}V_{a1}/ n_{a1}+ M_{a2}V_{a2}/ n_{a2}= M_{b}V_{b}/ n_{b}[(0.100 mol/L) (0.0300 L)] / 1 + [(x) (0.0200 L)] / 1 = [(0.200 mol/L) (0.0500 L)] / 1

0.003 + 0.02x = 0.01

0.02x = 0.007

x = 0.350 M

**Problem #10:** A flask contains 30.0 mL of 0.10 M hydrochloric and 20.0 mL of sulfuric acid. If 50.0 mL of 0.20 M sodium hydroxide was required to neutralize the mixture of acids in the flask, what was the concentration of the sulfuric acid?

**Solution:**

moles of NaOH ---> (0.20 mol/L) (0.0500 L) = 0.0100 molNaOH and HCl react in a 1:1 molar ratio. Let's see how much NaOH is left over after reacting with the HCl.

moles HCl ---> (0.10 mol/L) (0.0300 L) = 0.00300 mol

0.0100 mol minus 0.00300 mol = 0.00700 mol <--- left over NaOH

NaOH and H

_{2}SO_{4}react in a 2:1 molar ratio. To achieve neutralization, 0.00700 mol of NaOH reacts with 0.00350 mole of H_{2}SO_{4}.molarity ---> 0.00350 mol / 0.0200 L = 0.175 M

**Bonus Problem:** A mixture of hydrochloric acid and sulfuric acid is prepared so that it contains 0.635 M HCl and 0.135 M H_{2}SO_{4}. What volume of 0.140 M NaOH would be required to completely neutralize all of the acid in 588.8 mL of this solution?

**Solution:**

1) Determine the NaOH volume needed for the HCl:

HCl + NaOH ---> NaCl + H_{2}O(0.635 mol/L) (0.5888 L) = 0.373888 mol HCl present

By the above chemical equation, HCl and NaOH react in a 1:1 molar ratio. Therefore, 0.373888 mol of NaOH is required.

0.373888 mol / 0.140 mol/L = 2.67063 L

2) Determine the NaOH volume needed for the H_{2}SO_{4}:

H_{2}SO_{4}+ 2NaOH ---> Na_{2}SO_{4}+ 2H_{2}O(0.135 mol/L) (0.5888 L) = 0.079488 mol H

_{2}SO_{4}presentThe molar ratio between H

_{2}SO_{4}and NaOH is 1:2. Two moles of NaOH are required for every one mole of H_{2}SO_{4}(0.079488 mol) (2) = 0.158976 mol NaOH required

0.158976 mol / 0.140 mol/L = 1.13554 L

3) Add the two volumes:

2.67063 L + 1.13554 L = 3.80617 LTo three sig figs, the answer is 3.81 L.

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