100 ml of 0.1 M NaCl and 100 ml of 0.2 MNaOH are mixed. What is the change in pH of NaCI solution ?

100 ml of 0.1 M NaCl and 100 ml of 0.2 MNaOH are mixed. What is the ch

1.	100 ml of 0.1 M NaCl and 100 ml of 0.2 MNaOH are mixed. What is the change in pH of NaCI solution ?
Answer» <html><body><p> Some `KMnO_4`is <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> unreacted<br/>0.02 M `<a href="https://interviewquestions.tuteehub.com/tag/mn-548487" style="font-weight:bold;" target="_blank" title="Click to know more about MN">MN</a>^(2+)` ions are present in the solutions <br/>`K_2SO_3` is <a href="https://interviewquestions.tuteehub.com/tag/partially-7321885" style="font-weight:bold;" target="_blank" title="Click to know more about PARTIALLY">PARTIALLY</a> left unreacted <br/>Newly formed `SO_4^(2-)` has 0.05 M cocentration</p>Solution :`100xx0.2 =<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>` meq `O_3^(2-)` <br/> `100xx0.5 =50`meq `MnO_4^(-)` <br/> `:. SO_3^(2-)` in limiting <a href="https://interviewquestions.tuteehub.com/tag/reagent-1178480" style="font-weight:bold;" target="_blank" title="Click to know more about REAGENT">REAGENT</a> <br/> `MnO_4^(-)` left = 30 meq<br/> `:.[Mn^(+2)]=(20//5)/200=0.05M`<br/> `[SO_4^(-2)]=(20//2)/200 = 0.05M`</body></html>