Titration of Sr(OH)2 by HCl. What is the pH?

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SUMMARY

The discussion focuses on the titration of 80.0 mL of 0.100 M Sr(OH)2 with 0.400 M HCl, specifically calculating the pH at various volumes of HCl added. The initial pH before any HCl is added is calculated using the formula -log(0.1), yielding a pH of 13. The participants clarify that since Sr(OH)2 is a strong base and HCl is a strong acid, the Henderson-Hasselbalch equation is not applicable. The correct approach involves calculating the resulting pH based on the stoichiometry of the reaction: Sr(OH)2 + 2HCl → 2H2O + Sr.

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1. COnsider the titration of 80.0 mL of 0.100 M Sr(OH)2 by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.

a. 0.0 mL
b. 20 mL
c.80 mL

Homework Equations


Henderson–Hasselbalch equation or direct -log

The Attempt at a Solution


a) before I added any HCl, I just did the -log(.1)=1 14-1=13 is that correct?
B)i know it is a strong acid & strong base. & i cannot use pKa for Henderson–Hasselbalch equation. what else can i do?
Sr(OH)2 + 2HCl -> 2H2O + Sr
0.092 ... 0.0016 ... 0
-0.008 ... -0.0016 ... +0.008
0.0912 ... 0 ... . 0.008
 
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in this do i also have to multiply the HCL molarity by two or only the Sr(OH)2
for letter b i got a pH of 0 and when i did it with 30 mL i got a higher pH...is that normal?
 

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