Quick PH Question (1 Viewer)

[Zoinked]

2007 HSC Exam 21b) (No sample answers for some reason)

Using the red cabbage indicator, what colour would the solution be if 10 mL of
0.005 mol L–1 H2SO4 was diluted to 100 mL?

I did 0.005 x 2 x 0.1 = 0.001. Then -Log10(0.001) = Ph of 3 and that the indicator would turn violet.

Is that correctly done?

https://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/pdf_doc/chemistry_07.pdf

 

[leehuan]

a)
pH = -log₁₀(0.005*2)=2

This is enough information to say that your answer to part b) is a correct one.

(Naturally, we are assuming full ionisation of both protons of sulfuric acid, when in reality only one definitely ionises and the other just ionises noticeably easily)

Reason: A dilution by a factor of 10 is equivalent to decreasing the concentration 10-fold. A decrease in the concentration of the acid 10-fold also decreases the concentration of the hydronium ion by that much.

If the concentration of the hydronium ion is decreased 10-fold, we precisely increase the pH by one. This is because pH=-log₁₀[H_3OO⁺] is essentially a base 10 logarithmic scale.

 

[Zoinked]

Thanks leehuan, knew I could count on you