Question

calculate the concentration of an hcl solution if 100.0 ml of the hcl required 33.00 ml of 0.2000 m mg(oh)₂ to reach the titration endpoint.
mg(oh)₂ + 2 hcl → 2 h₂o + mgcl₂
there is enough information to calculate the moles of base but not the moles of acid.
1 mol
(2 l)(-----------) = 3 mol mg(oh)₃
1 l
4 mol hcl
(5 mol mg(oh)₃)(-------------------) = 6 mol hcl
7 mol mg(oh)₂
8 mol hcl
------------------- = 9 m hcl
10 l
a. 0.2000 b. 0.9960 c. 0.0040 d. al⁺³ e. co₂ f. caf₂(aq)
g. hf(aq) h. 0.03333 i. oh⁻¹ j. h₃o⁺¹ k. so₄⁻² l. mg⁺²
m. cl⁻¹ n. hpo₄⁻² o. h₂po₄⁻¹ p. co₃⁻² q. hco₃⁻¹ r. h₂s
s. hs⁻¹ t. s⁻² u. h₂o(liq) v. 6.600 x 10⁻³ w. 0.1320
x. 1.320 x 10⁻² y. 0.1000 z. 1 x 10⁻¹⁴ aa. 0.9923 bb. 0.0077
cc. 3 dd. 2 ee. 1 ff. 0.0154 gg. 3.95 x 10⁻³ hh. 1.84 x 10⁻⁶

Explanation:

Step1: Calculate moles of Mg(OH)₂

First, convert the volume of Mg(OH)₂ from mL to L. The volume is 33.00 mL, which is \( \frac{33.00}{1000} = 0.03300 \) L. The molarity of Mg(OH)₂ is 0.2000 M (which is 0.2000 mol/L). So moles of Mg(OH)₂ = volume (L) × molarity. So \( 0.03300 \, \text{L} \times \frac{0.2000 \, \text{mol}}{1 \, \text{L}} = 0.006600 \, \text{mol} \) (Wait, but let's check the options. Wait, maybe I made a mistake. Wait the options have V as \( 6.600 \times 10^{-3} \) which is 0.0066. Wait, but let's go step by step as per the given blanks.

Blank 2: volume of Mg(OH)₂ in L: 33.00 mL = 0.03300 L? Wait no, 33.00 mL is 0.03300 L? Wait 33 mL is 0.033 L? Wait 33.00 mL is 0.03300 L. But the options: Wait the first part: ( __2__ L)( __1__ mol / 1 L ) = __3__ mol Mg(OH)₂. Wait, molarity is mol/L, so the conversion factor is 0.2000 mol/L (option A). So blank 2: 33.00 mL = 0.03300 L? Wait but the options: Wait maybe the volume is 33.00 mL = 0.03300 L, but let's check the options. Wait the options for blank 2: maybe it's 0.03300 L, but the options have H as 0.03333? No, wait 33.00 mL is 0.03300 L. Wait maybe the problem has a typo, but let's proceed.

Wait the reaction is Mg(OH)₂ + 2 HCl → 2 H₂O + MgCl₂. So mole ratio of HCl to Mg(OH)₂ is 2:1.

So step 1: moles of Mg(OH)₂ = volume (L) × molarity. Volume is 33.00 mL = 0.03300 L (blank 2: 0.03300 L? But options: H is 0.03333, which is 33.33 mL? Wait no, 33.00 mL is 0.03300 L. Wait maybe the problem has 33.33 mL? No, the problem says 33.00 mL. Wait maybe I'm overcomplicating. Let's use the options.

Blank 2: volume of Mg(OH)₂ in L: 33.00 mL = 0.03300 L, but the options have H as 0.03333 (which is 33.33 mL). Wait maybe the problem has 33.33 mL? No, the problem says 33.00 mL. Wait maybe the first blank (blank 1) is the molarity, which is 0.2000 (option A). So blank 2: 0.03300 L? But options: H is 0.03333. Wait maybe the volume is 33.33 mL? No, the problem says 33.00 mL. Wait let's check the moles of Mg(OH)₂: 0.03300 L × 0.2000 mol/L = 0.006600 mol (which is \( 6.600 \times 10^{-3} \) mol, option V). So blank 3: V.

Then step 2: moles of HCl. Mole ratio is 2 mol HCl / 1 mol Mg(OH)₂ (option DD is 2). So moles of HCl = moles of Mg(OH)₂ × 2. So if moles of Mg(OH)₂ is 0.006600 mol (V), then moles of HCl = 0.006600 × 2 = 0.01320 mol (option W: 0.1320? No, 0.01320 is \( 1.320 \times 10^{-2} \), option X). Wait 0.0066 × 2 = 0.0132, which is \( 1.320 \times 10^{-2} \) (option X).

Then step 3: concentration of HCl. Volume of HCl is 100.0 mL = 0.1000 L (option Y: 0.1000). So concentration (M) = moles of HCl / volume (L) = 0.01320 mol / 0.1000 L = 0.1320 M (option W).

Wait let's fill the blanks:

Blank 2: volume of Mg(OH)₂ in L: 33.00 mL = 0.03300 L, but options have H as 0.03333. Wait maybe the problem has 33.33 mL? No, the problem says 33.00 mL. Wait maybe the first part:

( __2__ L)( __1__ mol / 1 L ) = __3__ mol Mg(OH)₂.

__1__ is molarity: 0.2000 mol/L (option A).

__2__ is volume: 33.00 mL = 0.03300 L, but option H is 0.03333. Wait maybe the volume is 33.33 mL? No, the problem says 33.00 mL. Wait maybe a mistake in the problem, but let's proceed with the options.

Assuming:

Blank 2: H (0.03333 L, which is 33.33 mL, maybe a typo).

Then moles of Mg(OH)₂: 0.03333 L × 0.2000 mol/L = 0.006666 mol (but option V is \( 6.600 \times 10^{-3} \), close).

Then mole ratio: 2 mol HCl / 1 mol Mg(OH)₂ (option DD: 2).

So moles of HCl: 0.006666 mol × 2 = 0.01333 mol (option X: \( 1.320 \times 10^{-2} \) is 0.0132, close).

Then volume of HCl: 100.0 mL = 0.1000 L (option Y).

Concentration: 0.0132 mol / 0.1000 L = 0.1320 M (option W).

So let's fill the blanks:

1: A (0.2000)

2: H (0.03333)

3: V (6.600 × 10⁻³)

4: DD (2)

5: V (6.600 × 10⁻³)

6: X (1.320 × 10⁻²)

7: EE (1)

8: X (1.320 × 10⁻²)

9: W (0.1320)

10: Y (0.1000)

But let's verify the calculation properly:

1. Moles of Mg(OH)₂: \( n = M \times V = 0.2000 \, \text{mol/L} \times 0.03300 \, \text{L} = 0.006600 \, \text{mol} \) (which is \( 6.600 \times 10^{-3} \) mol, option V).

2. Mole ratio from reaction: 2 mol HCl / 1 mol Mg(OH)₂ (so 4: DD (2), 7: EE (1)).

3. Moles of HCl: \( 0.006600 \, \text{mol} \times 2 = 0.01320 \, \text{mol} \) (option X: \( 1.320 \times 10^{-2} \) mol).

4. Volume of HCl: 100.0 mL = 0.1000 L (option Y).

5. Concentration of HCl: \( M = \frac{n}{V} = \frac{0.01320 \, \text{mol}}{0.1000 \, \text{L}} = 0.1320 \, \text{M} \) (option W).

So the blanks:

1: A

2: 0.03300 L (but option H is 0.03333, maybe a typo, but if we use 33.00 mL = 0.03300 L, but no option for that. Wait the options have H as 0.03333, which is 33.33 mL. Maybe the problem has 33.33 mL instead of 33.00 mL. Let's assume that:

Volume of Mg(OH)₂: 33.33 mL = 0.03333 L (option H).

Moles of Mg(OH)₂: \( 0.03333 \, \text{L} \times 0.2000 \, \text{mol/L} = 0.006666 \, \text{mol} \) (close to V: \( 6.600 \times 10^{-3} \), maybe rounding).

Then moles of HCl: \( 0.006666 \, \text{mol} \times 2 = 0.01333 \, \text{mol} \) (close to X: \( 1.320 \times 10^{-2} \), maybe the problem uses 33.00 mL).

Anyway, the final concentration of HCl is 0.1320 M (option W).

Answer:

The concentration of HCl solution is \(\boxed{0.1320}\) M (corresponding to option W).