plz help me with my chem questions (1 Viewer)

[lanvins]

Okay i'm gonna throw it out there. These are the question i can't do. Some of them r probably really easy but my textbook has no explanation

1. The concentration of sodium hyroxide in waste water from an alumina refinery was found by titrating 20ml aliquots of waste water agaisint .15M hydrochloric acid, using phenolphthalein as indciator. The average titre of several titration was 11.4ml. What mass of sodium hydroxide would be present in 100L of the waste water?

2. A 1.20 g anatacid tablet contains 80.% by mass of magnesium hyroxide as the acitve ingredient. What volume of .15M hydrochloric acid could the anatacid tablet neutralise?

3 What volumes of .1M sulfuric acid woould be requires to neutralise a solution containing .5g of sodium hydroxide and .8g of potassum hydroxide?

4. A certain brand of washing soda contains partially hydrated sodium carbonate solid. A .3g sample completely reacts with 20ml of .250M hydrochloric acid. Calculate the percentage by mass of sodium carbonate in the washing soda.

thanks a heapzzzzzzzzzzzzzzzzzzz

 

[undalay]

1.
Find number of moles used up in titration.
n = 0.15 x 11.4/1000
n = 1.71 x10^-3 to neutralise 20mls of unknown
5n = 8.55 x10^-3 to neutralise 100mls of unknown.
NaOH^- + HCl⁺ ----> H₂ O + NaCl(aq)
Thus 8.55 x10^-3 moles in 100mls of unknown.
8.55 x10^-3 x molar mass
8.55 x10^-3 x (22.99 +16 +1.008)
= 0.34 grams of sodium hydroxide in 100 mls (2sig figs)

2.
Mg(OH)₂ + 2HCl -> 2H₂ O + MgCl₂
1.2g of tablet.
thus .8x 1.2 of Mg(OH)₂ = 0.96 grams of Mg(OH)₂
MM_{Magnesium oxide} = 24.31+ 2(18+1.008) = 58.326g
m/MM = n
n(Magnesium oxide) = 0.01646
n(hydroxide) = 2xn(Magnesium oxide) = 0.03292
thus n(HCl) = n(hydroxide) = 0.03292

HCl: 0.15moles per 1L = 1.5x10^-4 per mL

n(HCl) / 1.5x10^-4 = number of mL
=220 mL (2 sig fig) of 0.15 M HCl solution.

3.
Sodium hydroxide = NaOH
MM_NaOH = (22.99+16+1.008) = 40
n = m/MM = 0.5 / 40 = 0.0125 moles
Potassium hydoxide = KOH
MM_NaOH = (39.10+16+1.008) = 56.1
n = m/MM = 0.8 / 56.1 = 0.01426

Total hydroxide = 0.0125+0.01426 = 0.2676 = moles of H+ needed.

H₂SO₄.
Thus n_H+/2 = n_H2SO4. = 0.2676 /2 = 0.01338

0.1moles per litre = 1x10^-4 per mL
n_H2SO4. /1x10^-4 = mL required
=130mL

 

[Mark576]

4. Na₂CO_3(aq)+2HCl_(aq) ---> H₂O_(l)+CO_2(g)+2NaCl_(aq)
Moles of HCl that reacted completely with the sodium carbonate present in the washing soda = 0.020 * 0.250 = 0.005 mol.
From the balanced equation we see that the moles of Na₂CO_3(aq) are then 0.5 * 0.005 = 0.0025 mol.
Therefore the mass of Na₂CO₃ = 0.0025 * [2(22.99)+48.00+12.01] = 0.26 g.
Hence percentage by mass of Na₂CO₃ in the sample of washing soda = 0.26/0.3 * 100 = 90% (1 sig fig)