2 What s the (1) equivalent weight, (2) # of equivalents, and (3) norm following

Question

2 What s the (1) equivalent weight, (2) # of equivalents, and (3) norm following? ality of each of the a. 0.800g NaOH in 500 ml (40.0g, 0.0200 equiv, 0.0400N) b. 18.5g Ba(OH)2 in 600 ml (85.5g, 0.216 equiv, 0.360N) 3. Calculate the normality of NaOH solutions which require: a. 42.0 ml to neutralize 22.0 ml of 0.200N HCI (0.105N) b. 21.8 ml to neutralize 26.0 ml of 0.750 N H2SO4 (0.894N) 4. What is the concentration of each of the following? a. 15.0 ml of 0.400N HCI diluted to 300 ml volume (0.0200N) b. 25.0ml of 3.00M H2S0, diluted to 100ml volume (0.750M) dissolved in 287ml of water and evaporated to 100ml final volume (2.00M)

Explanation / Answer

2. a. 0.800 g of NaOH in 500 ml volume:

1. Equivalent weight of a solution is the molar mass of the solute, or dissolved substance, in grams divided by the valence of the solute. To calculate Equivalent weight, we first calculate the molar mass and valnce of the compound.

Molar mass of NaOH = molar mass of Na (23.0) + mola mass of O(16.0) + molar mass of H(1.0) = 40.0 g/mol.

Valence of NaOH = 1, since one mole of OH- is produced per 1 mole of NaOH.

Equivalent weight = molar mass/valence = 40.0 g/mol/1 = 40.0 g/mol

2) No of equivalents:

1 equivalent weight (40.0 g) of NaOH would contain 1equivalent, so 0.800g would contain:

0.800g NaOH * (1 equivalent/40.0 g NaOH) = 0.0200 equivalents

3) Normality:

Molarity of NaOH = moles/volume in L

moles NaOH = mass/molar mass = 0.800g/40.0 g/mol = 0.0200 moles

Volume of NaOH = 500 ml = 0.5 L

Molarity of NaOH = 0.0200 moles/0.5 L = 0.04M.

Normality of NaOH = Molarity of NaOH*No of Equivalnets = 0.04*1 = 0.04N

2. a. 18.5 g of Ba(OH)2 in 600 ml volume:

1. Equivalent weight of a solution is the molar mass of the solute, or dissolved substance, in grams divided by the valence of the solute. To calculate Equivalent weight, we first calculate the molar mass and valnce of the compound.

Molar mass of Ba(OH)2 = molar mass of Ba (137.3) + 2*molar mass of O(16.0) + 2* molar mass of H(1.0) = 171.34 g/mol.

Valence of Ba(OH)2 = 2, since 2 moles of OH- is produced per 1 mole of Ba(OH)2.

Equivalent weight = molar mass/valence = 171.34 g/mol/2 = 85.67 g/mol

2) No of equivalents:

1 equivalent weight (171.34 g) of Ba(OH)2 would contain 2 equivalents, so 18.5 g would contain:

18.5 g Ba(OH)2 * (2 equivalent/171.34 g Ba(OH)2) = 0.216 equivalents

3) Normality:

Molarity of Ba(OH)2 = moles/volume in L

moles Ba(OH)2 = mass/molar mass = 18.5g/171.34 g/mol = 0.108 moles

Volume of Ba(OH)2 = 600 ml = 0.600 L

Molarity of NaOH = 0.108 moles/0.600 L = 0.180 M.

Normality of NaOH = Molarity of NaOH*No of Equivalents = 0.180 M * 2 = 0.360 N

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