A buffer containing 0.30 M HCLO and 0.30 M NaClO has a pH of 7.53. What will the pH be after .010 mol of NaOH is added to 100 ml of the buffer.
Answer options: 7.83
7.53
8.23
7.03
You want to use the HH equation (Henderson-Hasselbalch). The problem doesn't give you a pKa for HClO but you can calculate it.
pH = pKa + lob (base/acid)
7.53 = pKa + log (0.30/0.30)
pKa = 7.53
100 mL 0.30 M HClO = 30 millimoles HClO
100 mL 0.30 M NaClO = 30 millimiles NaClO
Add 0.010 mols NaOH = 10 millimoles NaOH added
..................HClO + OH^- ==> ClO^- + HOH
I...................30........0.................30................
add..........................10.....................................
C...............-10.........-10.............+10.......................
E..................20...........0.............+40....................
pH = pKa + log (base/acid)
pH = 7.53 + log (40/20). You can finish.
Technically, that 40/20 SHOULD BE (40/100/20/100). The 40/20 is the millimoles ratio. Technically concentration goes there which is 40/100 and 20/100 but I use millimoles almost exclusively since the volume ALWAYS WILL CANCEL. Post your work if you get stuck.
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To find the pH after adding NaOH to the buffer, we need to consider the reaction between NaOH and HCLO in the buffer solution.
First, let's write the balanced equation for the reaction:
HCLO + NaOH -> H2O + NaClO
Next, let's determine the initial moles of HCLO and NaClO before adding NaOH:
- Initial moles of HCLO = 0.30 M x 0.100 L = 0.030 moles (since the volume is given as 100 mL = 0.100 L)
- Initial moles of NaClO = 0.30 M x 0.100 L = 0.030 moles
When 0.010 moles of NaOH is added, it will react with the HCLO in a 1:1 ratio. Therefore, the moles of HCLO will decrease by 0.010 moles.
Now, let's find the final moles of HCLO and NaClO after the reaction:
- Final moles of HCLO = 0.030 moles - 0.010 moles = 0.020 moles
- Final moles of NaClO = 0.030 moles
To calculate the final concentration of HCLO and NaClO, divide the moles by the final volume of the solution. Since the volume remains the same (0.100 L), the concentrations of both HCLO and NaClO will be the same as the initial concentrations (0.30 M).
Since the initial pH of the buffer is given as 7.53, which indicates that the solution is slightly acidic, we know that HCLO is the dominant species. Therefore, the addition of NaOH will react with only HCLO and form water.
The pH of the solution will depend on the concentration of HCLO after the reaction. Since the concentration remains the same (0.30 M), the pH will remain the same as well.
Therefore, the correct answer is 7.53.
To determine the new pH after adding NaOH to the buffer solution, we need to consider the reaction that occurs between the added NaOH and the existing components of the buffer.
Firstly, let's write out the reaction that occurs when NaOH reacts with HCLO:
NaOH + HCLO → NaClO + H2O
In this reaction, NaOH reacts with HCLO to form NaClO and water.
Since NaOH is a strong base, it reacts completely with HCLO, leading to a decrease in the concentration of HCLO.
However, since NaClO is a weak base, it does not react completely with HCLO, resulting in an increased concentration of NaClO.
Now, let's calculate the new concentrations of HCLO and NaClO after adding 0.010 mol of NaOH to 100 mL of the buffer solution.
Initially, the buffer solution contains 0.30 M HCLO and 0.30 M NaClO. Adding 0.010 mol of NaOH to 100 mL of the buffer will lead to a total volume of 100 mL + 0.010 L = 0.110 L.
The moles of HCLO initially present in the buffer solution can be calculated as:
moles of HCLO = initial concentration of HCLO × volume of the solution
= 0.30 mol/L × 0.100 L
= 0.030 mol
The moles of NaClO initially present in the buffer solution can be calculated as:
moles of NaClO = initial concentration of NaClO × volume of the solution
= 0.30 mol/L × 0.100 L
= 0.030 mol
Now let's determine the final moles of HCLO and NaClO after the reaction with NaOH:
Since NaOH reacts with HCLO in a 1:1 ratio, the moles of NaOH reacted is equal to the moles of HCLO reacted, which is 0.010 mol.
After the reaction, the moles of HCLO remaining can be calculated as:
moles of HCLO remaining = initial moles of HCLO - moles of HCLO reacted
= 0.030 mol - 0.010 mol
= 0.020 mol
Similarly, the moles of NaClO formed can be calculated as:
moles of NaClO formed = initial moles of NaClO + moles of HCLO reacted
= 0.030 mol + 0.010 mol
= 0.040 mol
Next, we need to calculate the final concentrations of HCLO and NaClO.
Concentration of HCLO = moles of HCLO remaining / volume of the solution
= 0.020 mol / 0.110 L
= 0.182 M
Concentration of NaClO = moles of NaClO formed / volume of the solution
= 0.040 mol / 0.110 L
= 0.364 M
Finally, let's determine the new pH of the buffer solution using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
The pKa of HCLO is given as approximately 7.53 (since HCLO is a weak acid).
[A-] represents the concentration of NaClO, which is 0.364 M.
[HA] represents the concentration of HCLO, which is 0.182 M.
pH = 7.53 + log(0.364/0.182)
= 7.83
Therefore, the new pH of the buffer solution after adding 0.010 mol of NaOH is approximately 7.83.
Hence, the correct answer is 7.83.