Question

A strong acid-strong base titration was performed as follows: 10.00 ml of 0.10m hcl were collected in an erlenmeyer flask, and then diluted to 100 ml with distilled water. after that, two drops of indicator were added. then, the buret was filled with 0.100 m naoh. the titration was started by slowly adding naoh to hcl. (note: the volume of two drops indicator is ignored as it is so tiny compared with 100 ml.) what is the ph of the solution in the erlenmeyer flask after adding 2.00 ml of 0.100m naoh?

a. 3.2
b. 2.11
c. all options are incorrect
d. 1.22

Answer 1

The pH of the solution in the Erlenmeyer flask after adding 2.00 mL of 0.100 M NaOH is 7.00, making option c the correct answer.

Step-by-step explanation:

To calculate the pH of the solution in the Erlenmeyer flask after adding 2.00 mL of 0.100 M NaOH, we need to consider the reaction between HCl and NaOH. Both are strong acids and bases, respectively.

The balanced chemical equation for the reaction is:

HCl + NaOH → H2O + NaCl

This is a neutralization reaction, where the strong acid (HCl) reacts with the strong base (NaOH) to form water and salt. Since 1 mole of HCl reacts with 1 mole of NaOH, and the concentrations of both solutions are equal (0.100 M), the pH at the equivalence point is 7.00. Therefore, the correct option is c. All options are incorrect.

Answer 2

After adding 2.00 mL of 0.100 M NaOH to 10.00 mL of 0.10 M HCl, the excess HCl concentration in the final volume is approximately 0.0078 M. Taking the negative logarithm (base 10) of this concentration gives the pH, which is calculated to be around 2.11. Hence, the correct answer is (b). Thus the correct option is b. 2.11

Step-by-step explanation:

In a strong acid-strong base titration, the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be represented as follows:

`\[ HCl_((aq)) + NaOH_((aq)) \rightarrow H_2O_((l)) + NaCl_((aq)) \]`

Initially, we have 10.00 mL of (0.10 , M\) HCl, which is then diluted to 100 mL. The moles of HCl can be calculated as:

`\[ \text{moles} = \text{Molarity} * \text{Volume} \]`

`\[ \text{moles of HCl} = (0.10 \, M) * (0.010 \, L) = 0.001 \, mol \]`

Now, 2.00 mL of (0.100 , M) NaOH is added. The moles of NaOH added can be calculated similarly:

`\[ \text{moles of NaOH} = (0.100 \, M) * (0.002 \, L) = 0.0002 \, mol \]`

Since the balanced equation has a 1:1 ratio between HCl and NaOH, and the moles of NaOH added are less than the moles of HCl initially present, not all HCl will react. We need to find the excess moles of HCl:

`\[ \text{Excess moles of HCl} = \text{moles of HCl} - \text{moles of NaOH} \]`

`\[ \text{Excess moles of HCl} = 0.001 \, mol - 0.0002 \, mol = 0.0008 \, mol \]`

Now, we need to find the concentration of the excess HCl in the final volume (100 mL + 2 mL):

`\[ \text{New concentration of HCl} = \frac{\text{Excess moles of HCl}}{\text{Total volume}} \]`

`\[ \text{New concentration of HCl} = (0.0008 \, mol)/(0.102 \, L) \]`

\
`[ \text{New concentration of HCl} \approx 0.0078 \, M \]`

Taking the negative logarithm (base 10) of this concentration gives the pH:

`\[ \text{pH} = -\log(0.0078) \approx 2.11 \]`

Therefore, the pH of the solution after adding 2.00 mL of (0.100 , M\) NaOH is approximately 2.11, and the correct answer is b. 2.11