<u>Answer:</u> The
for calcium hydroxide is ![5.324\times 10^{-6}](https://tex.z-dn.net/?f=5.324%5Ctimes%2010%5E%7B-6%7D)
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![HCl](https://tex.z-dn.net/?f=HCl)
are the n-factor, molarity and volume of base which is ![Ca(OH)_2](https://tex.z-dn.net/?f=Ca%28OH%29_2)
We are given:
![n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL](https://tex.z-dn.net/?f=n_1%3D1%5C%5CM_1%3D0.0983M%5C%5CV_1%3D11.22mL%5C%5Cn_2%3D2%5C%5CM_2%3D%3FM%5C%5CV_2%3D50mL)
Putting values in above equation, we get:
![1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M](https://tex.z-dn.net/?f=1%5Ctimes%200.0983%5Ctimes%2011.22%3D2%5Ctimes%20M_2%5Ctimes%2050%5C%5C%5C%5CM_2%3D0.011M)
The concentration of
comes out to be 0.011 M.
The balanced equilibrium reaction for the ionization of calcium hydroxide follows:
![Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-](https://tex.z-dn.net/?f=Ca%28OH%29_2%5Crightleftharpoons%20Ca%5E%7B2%2B%7D%2B2OH%5E-)
The expression for solubility constant for this reaction follows:
![K_{sp}=[Ca^{2+}][OH^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BOH%5E-%5D%5E2)
Putting the values in above equation, we get:
![K_{sp}=(0.011)\times (2\times 0.11)^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%280.011%29%5Ctimes%20%282%5Ctimes%200.11%29%5E2)
![K_{sp}=5.324\times 10^{-6}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D5.324%5Ctimes%2010%5E%7B-6%7D)
Hence, the
for calcium hydroxide is ![5.324\times 10^{-6}](https://tex.z-dn.net/?f=5.324%5Ctimes%2010%5E%7B-6%7D)