Let's determine the concentration of a caesium hydroxide (CSOH) solution given the specifics of the problem.
### Step-by-Step Solution:
1. Identify Given Information:
- Volume of [tex]\( \text{HI} \)[/tex] solution: 257.0 mL
- Molarity of [tex]\( \text{HI} \)[/tex] solution: 1.02 M
- Volume of [tex]\( \text{CSOH} \)[/tex] solution: 130.0 mL
2. Calculate the Moles of [tex]\( \text{HI} \)[/tex]:
The molarity (M) of a solution is defined as the number of moles (n) of solute per liter (L) of solution. The formula to find the number of moles is:
[tex]\[
\text{moles} = \text{Molarity} \times \text{Volume (in liters)}
\][/tex]
Convert the volume of [tex]\( \text{HI} \)[/tex] from mL to L:
[tex]\[
257.0 \text{ mL} = 257.0 \div 1000 = 0.257 \text{ L}
\][/tex]
Now, calculate the moles of [tex]\( \text{HI} \)[/tex]:
[tex]\[
\text{moles of HI} = 1.02 \times 0.257 = 0.26214 \text{ moles}
\][/tex]
3. Understand the Reaction:
The neutralization reaction between [tex]\( \text{HI} \)[/tex] and [tex]\( \text{CSOH} \)[/tex] is given by:
[tex]\[
\text{HI} + \text{CSOH} \rightarrow \text{CsI} + \text{H}_2\text{O}
\][/tex]
From the equation, it's clear that 1 mole of [tex]\( \text{HI} \)[/tex] reacts with 1 mole of [tex]\( \text{CSOH} \)[/tex]. Therefore, the moles of [tex]\( \text{CSOH} \)[/tex] required will also be 0.26214 moles.
4. Calculate the Molarity of [tex]\( \text{CSOH} \)[/tex] Solution:
Using the volume of the [tex]\( \text{CSOH} \)[/tex] solution (130.0 mL), convert it from mL to L:
[tex]\[
130.0 \text{ mL} = 130.0 \div 1000 = 0.130 \text{ L}
\][/tex]
The molarity of the [tex]\( \text{CSOH} \)[/tex] solution is calculated using the formula:
[tex]\[
\text{Molarity} = \frac{\text{moles}}{\text{volume (in liters)}}
\][/tex]
Thus, the molarity of [tex]\( \text{CSOH} \)[/tex]:
[tex]\[
\text{Molarity of CSOH} = \frac{0.26214}{0.130} = 2.016 \text{ M}
\][/tex]
### Conclusion:
The concentration of the caesium hydroxide (CSOH) solution is 2.016 M.
