Can someone help me with this question please?
What is the pH at each of the points in the titration of 25.00 mL of 0.1000 MHCl by 0.1000 M NaOH:i) After adding 25.00 mL NaOH ii) After adding 26.00 mL NaOH
thankyou
This is a titration problem of a strong acid with a strong base.
HCl + NaOH ==> NaCl + H2O
These titrations can be divided into four distinct parts.
a. At the beginning. pH is determined by the pure HCl with no NaOH added.
b. At the equivalence point. pH is determined by the hydrolysis of the salt. IN this case, the salt is NaCl which is neutral so the pH is 7.
c. All points between a and b. moles HCl - moles NaOH = moles HCl remaining. pH = -log(H^+).
d. All points after b. All you have in the solution is excess NaOH. Total moles NaOH added - moles HCl initially = moles excess NaOH and M NaOH = moles NaOH/total volume. Then pOH = -log(OH^-) and pH + pOH = pKw 14.
Post your work if you get stuck.
To find the pH at each point in the titration, we need to understand the concept of titration and how it affects the pH. Titration is a process in which a solution of known concentration (titrant) is added to a solution of unknown concentration until a chemical reaction between the two solutions is complete.
In this case, we are titrating 25.00 mL of 0.1000 M HCl with 0.1000 M NaOH. This is an acid-base titration, where HCl is the acid and NaOH is the base. The reaction between HCl and NaOH is a neutralization reaction, resulting in the formation of water (H2O) and a salt (NaCl):
HCl + NaOH -> H2O + NaCl
To determine the pH after adding NaOH at each point, we need to calculate the moles of HCl and NaOH at each respective point.
i) After adding 25.00 mL NaOH:
In this case, we have added the same volume of NaOH as HCl. Therefore, the reaction between HCl and NaOH would be complete. Since we have equal moles of HCl and NaOH, they neutralize each other completely, resulting in a solution containing only the salt, NaCl.
ii) After adding 26.00 mL NaOH:
To determine the pH after adding an additional 1.00 mL of NaOH, we need to calculate the moles of HCl and NaOH at this point.
Moles of HCl = (volume of HCl in liters) x (molarity of HCl)
= (25.00 mL / 1000 mL/L) x (0.1000 M)
= 0.00250 moles HCl
Moles of NaOH = (volume of NaOH in liters) x (molarity of NaOH)
= (26.00 mL / 1000 mL/L) x (0.1000 M)
= 0.00260 moles NaOH
Since the reaction between HCl and NaOH is 1:1, the additional 0.00260 moles of NaOH would neutralize 0.00260 moles of HCl. Therefore, there would be a slight excess of NaOH in the solution.
To calculate the pH after adding the additional NaOH, we need to calculate the concentration of HCl and NaOH remaining in the solution.
Concentration of HCl = (moles of HCl remaining) / (volume of solution in liters)
= (0.00250 moles - 0.00260 moles) / (50.00 mL + 26.00 mL) / 1000 mL/L
= [(-0.00010) moles] / 0.076 L
= -0.001315 M (Note: The negative sign indicates an excess of NaOH)
Now, to calculate the pH, we can use the formula:
pH = -log[H+]
The concentration of H+ in the solution can be calculated using the concentration of HCl:
[H+] = [HCl] = 0.001315 M
Taking the negative logarithm:
pH = -log(0.001315) ≈ 2.88
In summary, after adding 25.00 mL NaOH, the solution would have a neutral pH (7). After adding 26.00 mL NaOH, the pH would be approximately 2.88.