acid base---TBR shortcut relevant

[unDRdog]

Given pH of CN- is 0.02M and has a Ka of 1.39 E-5. It ask you to calculate the pH... Can we use TBR shortcut of pH=1/2pKa -1/2log[HA]? This seems a little difficult because its hard for me to find 1/2log[log 0.002M].....
The answer is 10.7...anyone have an easy way of soving this w/o working the long [OH]^2/[CN]???

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[RedSoxSuck]

Given pH of CN- is 0.02M and has a Ka of 1.39 E-5. It ask you to calculate the pH... Can we use TBR shortcut of pH=1/2pKa -1/2log[HA]? This seems a little difficult because its hard for me to find 1/2log[log 0.002M].....
The answer is 10.7...anyone have an easy way of soving this w/o working the long [OH]^2/[CN]???

Is is the concentration 0.02M or 0.002M?

 

[unDRdog]

0.02M This question is from a GS test...

 

[RedSoxSuck]

0.02M This question is from a GS test...

Not sure if this is the right way to do it. So CN- is a base and they gave you ka.

Ka= 1.39 x 10^-5

Pka= 5-log 2 (i approximated 1.39 to 2 and this will probably e(a)ffect the answer.

pka = 5 - .3 = ~4.7

1/2pka +1 (since concentration of CN is 0.02) = 1/2 (4.7) + 1 = 3.35

14 - 3.35 = 10.65....


Not sure if thats the correct way..lol

 

[unDRdog]

Not sure if this is the right way to do it. So CN- is a base and they gave you ka.

Ka= 1.39 x 10^-5

Pka= 5-log 2 (i approximated 1.39 to 2 and this will probably e(a)ffect the answer.

pka = 5 - .3 = ~4.7
+1 (since concentration of CN is 0.02) = 1/2 (4.7) + 1 = 3.35

1/2pka 14 - 3.35 = 10.65....


Not sure if thats the correct way..lol

Sweet..can you explain the approximation of 1 please...now i'm feeling stuipid

 

[RedSoxSuck]

Sweet..can you explain the approximation of 1 please...now i'm feeling stuipid

well, TBR gen chem has a shortcut which you explained and used. 1/2pka.

It also says that if the initial concentration is 1.00 then it is just 1/2pka

If the concentration is 0.10 then it is 1/2pka + 0.5

If the concentration is 0.010 then it is 1/2pka + 1.00

In this question concentration was 0.02 which is close enough to 0.010, i added 1.0 to the 1/2pka.



EDIT: you add 1.0 and 0.5 when dealing with acid or base

 

[unDRdog]

well, TBR gen chem has a shortcut which you explained and used. 1/2pka.

It also says that if the initial concentration is 1.00 then it is just 1/2pka

If the concentration is 0.10 then it is 1/2pka + 0.5

If the concentration is 0.010 then it is 1/2pka + 1.00

In this question concentration was 0.02 which is close enough to 0.010, i added 1.0 to the 1/2pka.



EDIT: you add 1.0 and 0.5 when dealing with acid or base

Sweet!!! that helps alot!!! some how i missed that in the gen chem book...thanks again...you da man!

 

[Tokspor]

Given pH of CN- is 0.02M and has a Ka of 1.39 E-5. It ask you to calculate the pH... Can we use TBR shortcut of pH=1/2pKa -1/2log[HA]? This seems a little difficult because its hard for me to find 1/2log[log 0.002M].....
The answer is 10.7...anyone have an easy way of soving this w/o working the long [OH]^2/[CN]???

Would my way of doing this problem be correct? I got 8.5 at the end instead of 10.7 so I'm not sure.

We know [CN-] and K(a), with K(a) seemingly for HCN.

K(a)*K(b) = K(w)
(1.39 x 10E-5)*K(b) = 1.0 x 10E-14
K(b) = 7.5 x 10E-10

K(b) = [OH-][HCN]/[CN-]
7.5 x 10E-10 = x^2/(1.0 x 10E-2)
x = 4 x 10E-6

[OH-] = 4 x 10E-6
pOH = 5.5

pH = 14 - 5.5 = 8.5

Am I doing something wrong here?

 

[Isoprop]

You made a small mistake, but your final answer is close to mine (8.5). when i calculated with the Kb = 1.39 E-5, I get 10.7.

I think the OP meant Kb instead of Ka because looking up the pKa of HCN, it's 9.7.

Unless I missed something entirely.

Edit: your small mistake was that you used [HCN] = 0.01 M instead of 0.02.

 

[Tokspor]

You made a small mistake, but your final answer is close to mine (8.5). when i calculated with the Kb = 1.39 E-5, I get 10.7.

I think the OP meant Kb instead of Ka because looking up the pKa of HCN, it's 9.7.

Unless I missed something entirely.

Edit: your small mistake was that you used [HCN] = 0.01 M instead of 0.02.

Ah, thank you.

 

[BerkReviewTeach]

Would my way of doing this problem be correct? I got 8.5 at the end instead of 10.7 so I'm not sure.

We know [CN-] and K(a), with K(a) seemingly for HCN.

K(a)*K(b) = K(w)
(1.39 x 10E-5)*K(b) = 1.0 x 10E-14
K(b) = 7.5 x 10E-10

K(b) = [OH-][HCN]/[CN-]
7.5 x 10E-10 = x^2/(1.0 x 10E-2)
x = 4 x 10E-6

[OH-] = 4 x 10E-6
pOH = 5.5

pH = 14 - 5.5 = 8.5

Am I doing something wrong here?

I added a step to show where I think the error occured.

K(b) = [OH-][HCN]/[CN-]
7.5 x 10E-10 = x^2/(1.0 x 10E-2)
(7.5 x 10E-10) x (1.0 x 10E+2) = x^2
7.5 x 10E-8 = x^2
x = 3 x 10E-4

[OH-] = 3 x 10E-4
pOH = 4 - log 3 = 4 - 0.48 = 3.5ish

pH = 14 - 3.5ish = 10.5ish

But why do the long winded path when RedSoxSuck has shown a perfect short cut used EXACTLY the way it should be used?

well, TBR gen chem has a shortcut which you explained and used. 1/2pka.

It also says that if the initial concentration is 1.00 then it is just 1/2pka

If the concentration is 0.10 then it is 1/2pka + 0.5

If the concentration is 0.010 then it is 1/2pka + 1.00

In this question concentration was 0.02 which is close enough to 0.010, i added 1.0 to the 1/2pka.

EDIT: you add 1.0 and 0.5 when dealing with acid or base

This addition of 0.5 or 1.0 (etc...) can be modified and bulit into the equation as follows:

pH = (pKa + pHif it were a strong acid and not a weak acid)/2

In this case, it would be:
pOH = (pKb + pHif it were a strong base and not a weak base)/2

A strong base with a concentration of 0.02 M would be 2 - log 2 = 1.7. I believe that the Ka for HCN is 7.5 x 10exp-10 and that Kb for CN- is 1.3 x 10exp-5. That means that pKb is 5 - log 1.3 = 4.9

• So, the pOH would be:
  pOH = (4.9 + 1.7)/2 = 6.6/2 = 3.3
  
  pH = 14 - pOH = 14 - 3.3 = 10.7

It shgould be a matter of getting used to logs. If you used gthe short cut as Red

 

[BerkReviewTeach]

Would my way of doing this problem be correct? I got 8.5 at the end instead of 10.7 so I'm not sure.

We know [CN-] and K(a), with K(a) seemingly for HCN.

K(a)*K(b) = K(w)
(1.39 x 10E-5)*K(b) = 1.0 x 10E-14
K(b) = 7.5 x 10E-10

K(b) = [OH-][HCN]/[CN-]
7.5 x 10E-10 = x^2/(1.0 x 10E-2)
x = 4 x 10E-6

[OH-] = 4 x 10E-6
pOH = 5.5

pH = 14 - 5.5 = 8.5

Am I doing something wrong here?

I think the error is that Ka for HCN is 7.5 x 10exp-10, meaning that Kb is 1.3 x 10exp-5..

K(b) = [OH-][HCN]/[CN-]
1.3 x 10E-5 = x^2/(2.0 x 10E-2)
2.6 x 10E-7 = x^2
26 x 10E-8 = x^2
x = 5 x 10E-4

[OH-] = 5 x 10E-4
pOH = 4 - log 5 = 4 - 0.7 = 3.3ish

pH = 14 - 3.3ish = 10.3ish

But why do the long winded path when RedSoxSuck has shown a perfect short cut used EXACTLY the way it should be used?

well, TBR gen chem has a shortcut which you explained and used. 1/2pka.

It also says that if the initial concentration is 1.00 then it is just 1/2pka

If the concentration is 0.10 then it is 1/2pka + 0.5

If the concentration is 0.010 then it is 1/2pka + 1.00

In this question concentration was 0.02 which is close enough to 0.010, i added 1.0 to the 1/2pka.

EDIT: you add 1.0 and 0.5 when dealing with acid or base

• If you used the short cut as RedSoxSuck explained it, then you'd get essentially the same answer, but faster. Half of 4.9 is about 2.5. Adding 1.0 leads to a pOH of 3.5. If pOH is 3.5, then pH is 10.5.

This addition of 0.5 or 1.0 (etc...) approach can be modified and bulit into the equation as follows:

pH = (pKa + pHif it were a strong acid and not a weak acid)/2

In this case, it would be:
pOH = (pKb + pHif it were a strong base and not a weak base)/2

A strong base with a concentration of 0.02 M would be 2 - log 2 = 1.7. I believe that the Ka for HCN is 7.5 x 10exp-10 and that Kb for CN- is 1.3 x 10exp-5. That means that pKb is 5 - log 1.3 = 4.9

• So, the pOH would be:
  pOH = (4.9 + 1.7)/2 = 6.6/2 = 3.3
  
  pH = 14 - pOH = 14 - 3.3 = 10.7

This alternate method is also a TBR shortcut for those who feel better being a little more precise than the +0.5, +1.0, etc... method.

 

[evusq]

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