pH QUESTION! (1 Viewer)

[kingkong123]

Hey guys, could u have a look at this question?

<a href="http://www.codecogs.com/eqnedit.php?latex=\\$A small amount of pure sodium metal is dropped ino 1.2L of water. The reaction is summaries in the following equation: $\\\\2\mathbf{Na_{(s)}@plus;2H_2O_{(l)}\rightarrow2NaOH_{(aq)}@plus;H_{2(g)}}\\\\$ If the volume of gas collected occupied a volume of 4.68L at 25$^{o}$C and 100kPa, Calculate the final pH of the water.$\\\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\$A small amount of pure sodium metal is dropped ino 1.2L of water. The reaction is summaries in the following equation: $\\\\2\mathbf{Na_{(s)}+2H_2O_{(l)}\rightarrow2NaOH_{(aq)}+H_{2(g)}}\\\\$ If the volume of gas collected occupied a volume of 4.68L at 25$^{o}$C and 100kPa, Calculate the final pH of the water.$\\\\" title="\\$A small amount of pure sodium metal is dropped ino 1.2L of water. The reaction is summaries in the following equation: $\\\\2\mathbf{Na_{(s)}+2H_2O_{(l)}\rightarrow2NaOH_{(aq)}+H_{2(g)}}\\\\$ If the volume of gas collected occupied a volume of 4.68L at 25$^{o}$C and 100kPa, Calculate the final pH of the water.$\\\\" /></a>

I calculated an answer to be 13.51 but im not sure if its right. I can outline my working if needed. Can someone confirm the answer or outline a solution?

Thanks

 

[deswa1]

I got 13.50, which is probably the same answer- you might have rounded at some point or something.

 

[Aesytic]

i got 13.50 as well

 

[kingkong123]

thanks alot! could u guys have a look at this question as well. im not sure of the process

<a href="http://www.codecogs.com/eqnedit.php?latex=\\$0.52g of calcium carbonate was dissolved in 35mL of HCl of unknown molarity. After all the reaction had ceased and any CO$_2$ was removed by boiling, the sample was cooled. Methyl red was added and the solution was titrated with 0.1M NaOH. 22.3mL was needed to reach the endpoint. Find the molarity of HCl$." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\$0.52g of calcium carbonate was dissolved in 35mL of HCl of unknown molarity. After all the reaction had ceased and any CO$_2$ was removed by boiling, the sample was cooled. Methyl red was added and the solution was titrated with 0.1M NaOH. 22.3mL was needed to reach the endpoint. Find the molarity of HCl$." title="\\$0.52g of calcium carbonate was dissolved in 35mL of HCl of unknown molarity. After all the reaction had ceased and any CO$_2$ was removed by boiling, the sample was cooled. Methyl red was added and the solution was titrated with 0.1M NaOH. 22.3mL was needed to reach the endpoint. Find the molarity of HCl$." /></a>

thanks!

 

[kingkong123]

?

 

[Aesytic]

i think firstly you find how many moles of HCl are used up in the reaction between the CaCO3 and HCl by finding the number of moles of CaCO3 then using mole ratios. after that, you find the number of moles of NaOH that were used to neutralise the excess HCl, then use mole ratios again to find the number of moles of excess HCl. add both values to find the total number of moles of HCl then use n=cV to find molarity

 

[InkedFeather]

I got pH=13.57 but I didn't round at all so that might explain the difference.

 

[someth1ng]

13.50 is the correct answer - the calculator is 13.49716...but that rounds to 13.50.

 

[Octagon]

LOL i got 13.57, didnt round till the end, dunno where I went wrong...Did you guys go pH = 14 - pOH?

 

[nightweaver066]

LOL i got 13.57, didnt round till the end, dunno where I went wrong...Did you guys go pH = 14 - pOH?

You get that if you just find the moles of NaOH produced and then using pOH = -log[OH] immediately.

What you need to take in to account is the decrease in H2O because it reacts, then finding the new volume, and then the [OH-] and proceed from there.

 

[Chicharito]

Can someone please show me their working out for this question

 

[nightweaver066]

Can someone please show me their working out for this question

Moles of H2 produced = 4.68/24.79 = .1888
n(H2):n(H2O) = 1:2
.'. n(H2O) used = .1888 x 2 = .3776
Volume remaining = 1200 - .3776 x (1.008 x 2 + 16)
= 1193.2mL = 1.1932L
n(H2):n(NaOH) = 1:2
n(NaOH) = 0.3776
.'. n(OH-) = 0.3776
[OH-] = 0.3776/1.1932 = .31644
pOH = -log(.31644) = .49971
pH + pOH = 14
pH = 14 - pOH
= 13.50

 

[deswa4]

thanks alot! could u guys have a look at this question as well. im not sure of the process

<a href="http://www.codecogs.com/eqnedit.php?latex=\\$0.52g of calcium carbonate was dissolved in 35mL of HCl of unknown molarity. After all the reaction had ceased and any CO$_2$ was removed by boiling, the sample was cooled. Methyl red was added and the solution was titrated with 0.1M NaOH. 22.3mL was needed to reach the endpoint. Find the molarity of HCl$." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\$0.52g of calcium carbonate was dissolved in 35mL of HCl of unknown molarity. After all the reaction had ceased and any CO$_2$ was removed by boiling, the sample was cooled. Methyl red was added and the solution was titrated with 0.1M NaOH. 22.3mL was needed to reach the endpoint. Find the molarity of HCl$." title="\\$0.52g of calcium carbonate was dissolved in 35mL of HCl of unknown molarity. After all the reaction had ceased and any CO$_2$ was removed by boiling, the sample was cooled. Methyl red was added and the solution was titrated with 0.1M NaOH. 22.3mL was needed to reach the endpoint. Find the molarity of HCl$." /></a>

thanks!

What did u guys get for this question?
I got pH (HCl) = 0.632
Dunno if it's right :S