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The image you uploaded contains a titration problem for calculating the concentration of an HCl solution. Here’s a summary and solution approach based on the content visible:

**Problem Overview:**
- A titration was performed, and 19.90 mL of 0.157 M NaOH was needed to neutralize 25.0 mL of HCl.
- The balanced chemical equation is:
  \[
  \text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}
  \]
  This indicates a 1:1 molar ratio between HCl and NaOH.

**Formula to Use:**
- The problem suggests using the titration formula:
  \[
  M_A V_A = M_B V_B
  \]
  where:
  - \( M_A \) is the molarity of HCl (unknown),
  - \( V_A \) is the volume of HCl (25.0 mL),
  - \( M_B \) is the molarity of NaOH (0.157 M),
  - \( V_B \) is the volume of NaOH (19.90 mL).

**Solution Steps:**
1. Rearrange the formula to solve for \( M_A \):
   \[
   M_A = \frac{M_B V_B}{V_A}
   \]
2. Substitute the values:
   \[
   M_A = \frac{0.157 \times 19.90}{25.0}
   \]
3. Perform the calculation:
   \[
   M_A = \frac{3.1243}{25.0} = 0.12497 \, \text{M} \approx 0.125 \, \text{M}
   \]

**Answer:**
The concentration of the HCl solution is approximately **0.125 M**.

Would you like more detailed steps or explanations? Here are 5 related questions you might explore:

1. What are the units for molarity and why are they important?
2. How does the 1:1 molar ratio between HCl and NaOH affect the calculation?
3. Could this method be used if the molar ratio were not 1:1?
4. What is the significance of the balanced chemical equation in titration problems?
5. How would you handle experimental errors in a real titration process?

**Tip:** Always ensure volumes are in the same units when applying the titration formula.

By titration, 19.90 mL of 0.157 M NaOH is needed to neutralize 25.0 mL of HCl. Calculate the concentration of the HCl solution.

Solve a titration problem where 19.90 mL of 0.157 M NaOH neutralizes 25.0 mL of HCl. Learn to calculate the concentration of HCl using the formula M_A V_A = M_B V_B. Suitable for students in Grades 10-12.

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