Q66E
Question
The equivalence point for the titration of a \(25.00 - mL\)sample of \(CSOH\) solution with \(0.1062MHN{O_3}\) is at\(35.27mL\). What is the concentration of the \(CsOH\)solution?
Step-by-Step Solution
VerifiedThe CsOH has a concentration of 0.150M.
- Caesium hydroxide, also known as cesium hydroxide, is a chemical compound made up of caesium ions and hydroxide ions.
- It has a strong base (pKa= 15.76), similar to the other alkali metal hydroxides such as sodium and potassium.
The concentration of CsOH solution must be estimated using the equivalency point observed after titrating CsOH with \(HN{O_3}\)
The equivalency point (or volume) in an acid-base titration is the point (or volume) at which the titrant entirely neutralises the analyte.
At the point of equivalence:
Number of acid moles Equals Number of base moles
Given
Concentration of \(HN{O_3} = 0.1062M\)
Volume of \(HN{O_3}\) added to reach end point \( = 35.27ml = 0.03527\;L\)
Volume of CsOH taken \( = 25.00ml = 0.025\;L\)
The response of \(HN{O_3}\)with CsOH can be expressed as follows:
\(HN{O_3} + CsOH \to CsN{O_3} + {H_2}O\)
Based on the stoichiometry of the reaction:
1 mole of \(HN{O_3}\) 1 mole of a substance is combined with 1 mole of a substance CsOH
At the point of equivalence:
Moles of \(HN{O_3}\)consumed = Moles of CsOH present
Moles of \(HN{O_3}\) consumed \( = V\left( {HN{O_3}} \right) \times M\left( {HN{O_3}} \right)\)
\( = 0.03527\;L \times 0.1062\) moles. \({L^{ - 1}} = 0.00375\)moles
Moles of \(CsOH = 0.00375 = V(CsOH) \times M(CsOH)\)
\(M(CsOH) = \frac{{0.00375 moles }}{{0.025\;L}} = 0.15M\)
CsOH has a concentration of 0.150M.