Calculate the pH at the point in the titration of 25.00 mL 0.108 M HNO2 at which 10.00 mL 0.162 M NaOH have been added. For HNO2, Ka=5.1 x 10^-4 and:
HNO2 + OH- -----> H2O + NO2-
I know that HNO2 is a weak acid and NaOH is a strong base.
Help given to me so far from Dr. Bob222:
HNO2 + NaOH ==> NaNO2 + H2O
moles HNO2 = M x L = 0.025 x 0.108 = 0.00270 moles HNO2 initially.
moles NaOH added = 10mL x 0.162 = 0.00162
moles HNO2 remaining after reaction = 0.00270-0.00162 =0.00108.
M HNO2 = moles/L (L = 25 mL + 10 mL)
So this problem becomes determine the pH of a 0.00108/0.035)M soln of HNO2. Set up an ICE chart, solve for H^+ and convert to pH……
I set up the ICE chart:
HNO2 + H20 H3O^+ + NO2^- Ka=5.1 x10^-4
I: 3.085x10^-2 M --- ----
C: -X +X +X
E: 3.085x10^-2 –X M XM XM
Since the Ka and the Molarity of the acid are within 10^2, I cannot drop the –X. So, 5.1x10^-4= [X^2]/[3.085x10^-2 –X]. I get a quadratic equation of X^2+5.1x10^-4X -1.5734x10^-5=0. Therefore X=[H3O^+]=0.003719. pH=-log [X]. pH=2.43. My book says the answer is 3.47. I am confused. Please tell me what I am doing wrong. Thank you!
First, I divided 2.70 mmoles (0.00270 moles) by 35 = 0.03086M which isn't enough difference to worry about. Besides we don't have that many sig digits, anyway. The main thing is that you didn't include the NaNO2 that is formed. That is the salt of the weak acid so it is a buffered solution. (NaNO2) = 2.70mmoles/35mL = 0.0771M
Ka = 5.1E-4 = ((H^)(NO2^-)/(HNO2)
5.1E-4 = (x)(x+0.0771)/(0.03086-x).
I worked it both with a quadratic and without and it doesn't make much difference; however, I did NOT obtain 3.47. I found 2.02E-4 for (H^+) for pH = 3.69. I also used the Henderson-Hasselbalch equation which is
pH = pKa + log (NO2^-)/(HNO2)
pH = 3.29 + log(2.70mmols/1.08 mmols)
pH = 3.687 which would round to 3.69
I don't believe the 3.47 answer is correct. Check my work carefully.
I got it! You took the wrong mol of No2- from the above neutralization rxn....it should be 0.00162 mol NO2-/0.035L=0.0463M NO2-....I did all the calcs and I get 3.47 for pH which is book answer! Thank you for pointing me in the right direction! :-)
In your setup of the ICE table for the reaction of HNO2 with water, you missed a crucial step. When HNO2 reacts with water, it also produces H2O as a product. Therefore, the reaction should be written as:
HNO2 + H2O ⇌ H3O+ + NO2-
Now, let's re-calculate the concentrations using the correct setup:
Initial moles of HNO2 = M(HNO2) x V(HNO2) = 0.108 M x 0.025 L = 0.0027 moles
Moles of NaOH added = M(NaOH) x V(NaOH) = 0.162 M x 0.01 L = 0.00162 moles
Moles of HNO2 remaining after the reaction = Initial moles - Moles of NaOH added = 0.0027 - 0.00162 = 0.00108 moles
Now, to calculate the concentration of H3O+, divide the moles of HNO2 remaining by the total volume of the solution:
M(HNO2) = Moles HNO2 remaining / Total volume = 0.00108 moles / (0.025 L + 0.01 L) = 0.00108 moles / 0.035 L = 0.03086 M
Since the Ka is small, we can assume that the concentration of H3O+ formed is negligible compared to the initial concentration of HNO2. Therefore, we can approximate the concentration of H3O+ to be equal to x (assumed).
Using the expression for the dissociation constant (Ka = [H3O+][NO2-] / [HNO2]), we can substitute the values:
5.1 x 10^-4 = x^2 / (0.03086 - x)
Solve the equation for x using the quadratic formula, and you will find x = 0.003346 M, which is the concentration of H3O+.
Now, calculate the pH using the equation pH = -log[H3O+]:
pH = -log(0.003346) = 2.476
Therefore, the correct answer for the pH at the point in the titration is approximately 2.48, not 3.47 as stated in your book.
To find the pH at the point in the titration where 10.00 mL of 0.162 M NaOH has been added to 25.00 mL of 0.108 M HNO2, you need to consider the reaction between HNO2 and NaOH. From the balanced equation:
HNO2 + OH- → H2O + NO2-
You correctly determined the initial moles of HNO2 to be 0.00270 moles (0.025 L x 0.108 M).
The moles of NaOH added is 0.00162 moles (0.010 L x 0.162 M).
To find the moles of HNO2 remaining after the reaction, subtract the moles of NaOH added from the initial moles of HNO2:
moles HNO2 remaining = 0.00270 - 0.00162 = 0.00108 moles.
Now, you need to calculate the molarity of HNO2 in the final solution. The final volume is the sum of the initial volumes of HNO2 and NaOH:
Final volume = 25.00 mL + 10.00 mL = 35.00 mL = 0.035 L.
Molarity HNO2 = moles HNO2 / final volume = 0.00108 moles / 0.035 L = 0.0309 M.
So, the problem reduces to calculating the pH of a 0.0309 M solution of HNO2.
To solve for the pH, first, set up an ICE (Initial, Change, Equilibrium) table for the dissociation of HNO2:
HNO2 + H2O → H3O+ + NO2-
I: 0.0309 M 0 0
C: -x +x +x
E: 0.0309 - x x x
Note that since the initial concentration is relatively large (0.0309 M), you can assume that the change in concentration (-x) is negligible compared to the initial concentration.
Now, you can write the expression for the Ka of HNO2:
Ka = [H3O+][NO2-] / [HNO2]
You correctly set up the equation:
5.1 x 10^-4 = (x)(x) / (0.0309 - x)
Solving this quadratic equation, you found that x, which represents the concentration of H3O+, is 0.003719 M.
To find the pH from the H3O+ concentration, take the negative logarithm:
pH = -log[H3O+] = -log(0.003719) = 2.43.
However, your book states that the answer should be 3.47, which indicates that there may be an error in your calculations.
One possible error could be in the setup of the ICE table. Double-check that you correctly subtracted the change in concentration (-x) from the initial concentration (0.0309 M) to get the equilibrium concentration (0.0309 - x) in the E row.
Also, make sure you performed the quadratic equation correctly. It may be helpful to use an online quadratic equation solver or a graphing calculator to verify your solution.
Check your calculations step by step to see if there are any other errors. With the correct calculations, you should be able to obtain the correct pH value.