Author Topic: A few more questions..... (Read 845 times) Tweet Share

zibb3r · « **on:** February 03, 2011, 08:08:30 pm »

Guys, if I'm being annoying please stop me! I didn't do 1 and 2 chem and I have been trying to do the 1/2 course over the holidays. Thus, the hoard of questions!

I got really great previous answers and you guys have helped me understand a lot!!!! Thanks! :smitten:

So if you guys are up to it

:

Concentration:

1) A 12ml sample of a 15M solution of HCl is diuted to 1L, what is the final concentration of the acid.

2)4.5g of sodium chloride is added to 150ml of water in a volumetric flask. Tis solution is diluted to a final volume of 800ml. Find the concentration of the final solution.

Stoichiometry:

1) How many grams of sodium hydroxide can be produced from 500g of calcium hydroxide according to this equation:

Ca(OH)₂ + Na₂CO₃--->2NaOH + CaCO₃

pH:

1) If 3g of NaOH is added to 500ml of 0.10M HCl will the resulting solution be acidic or basic? What will the pH be???
THanks soo sooooooo sooooooooooo very much :angel: :smitten:

m@tty · « **Reply #1 on:** February 03, 2011, 08:18:17 pm »

1.

$C_1V_1=C_2V_2$ (volume in litres and concentration in M)

$C_2=\frac{0.012\times 15}{1}=0.18\text{ M}$

2.

$n(NaCl)=\frac{4.5}{23.0+35.5}=0.769\text{ mol}$

$C=\frac{n}{V}\text{ M}$

$\therefore C(NaCl)=\frac{0.77}{0.150}=5.128\text{ M}$

Again, $C_1V_1=C_2V_2$

$C_2=\frac{5.128 \times 0.150}{0.800}=0.97\text{ M}$ (two significant figures)

m@tty · « **Reply #2 on:** February 03, 2011, 08:20:16 pm »

A better, faster way with 2. is to just take the final volume because in the end you just have 4.5 g NaCl in 800mL of water.

So, $C(NaCl)=\frac{0.0769}{0.800}=0.97\text{ M}$ (2 sig figs)

m@tty · « **Reply #3 on:** February 03, 2011, 08:26:02 pm »

1) How many grams of sodium hydroxide can be produced from 500g of calcium hydroxide according to this equation:

Ca(OH)2 + Na2CO3--->2NaOH + CaCO3

$n(Ca(OH)_2)=\frac{500}{40+2(17)}=6.757\text{ mol}$

Mole ratio Ca(OH)2 : NaOH = 1:2

Therefore $n(NaOH)=2n(Ca(OH)_2)=13.514\text{ mol}$

And $m(NaOH)=13.514\times (23+17)=540.54=540\text{ g}$ (3 sig figs)

m@tty · « **Reply #4 on:** February 03, 2011, 08:35:54 pm »

pH:

1) If 3g of NaOH is added to 500ml of 0.10M HCl will the resulting solution be acidic or basic? What will the pH be???

$n(NaOH)=\frac{3}{40}=n(OH^-)=0.075\text{ mol}$

$n(HCl)=n(H^+)=cV=0.10 \times 0.500 = 0.05\text{ mol}$

When mixed you get a reaction between the OH- and H+ and there will be an excess of one reactant. This excess determines the pH.

Clearly there is 0.025 mole more OH-. So the solution will be basic.

So $[OH^-]=\frac{0.025}{0.5}=0.5M$

Now, in this situation, I figure out the pOH and subtract that from 14 (as at standard laboratory conditions (SLC - 298K, 1atm etc.) pH+pOH=14.)

So $\text{pOH}=-\log_{10}(0.5)=0.30$

So $\text{pH}=14-0.30=13.7$

pi · « **Reply #5 on:** February 04, 2011, 01:35:46 pm »

Another quick trick:

If your at SLC, pH = 14 + log₁₀[OH^-]

Saves seconds/minutes (depends how practiced you are) off those last three steps!

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