Question

Consider the following indicators and their pH ranges:

Answer 1

Answer:

c) 0.100M NH3(Kb=1.8×10^-5)+0.100M HCl.

Explanation:

Hello,

In this case a pH of about 5 is what we should look for, thus, we compute the pH of a 0.100M solution of ammonia as:

$NH_3NH_4^++OH^-\\K_b=\frac{[NH_4^+][OH^-]}{[NH_3]} \\K_b=\frac{x^2}{0.100M-x} \\0.100K_b-K_bx-x^2=0\\1.8x^{-6}-1.8x^{-5}x-x^2=0\\x=1.3x10^{-3}M\\[OH^-]=1.3x10^{-3}\\pH=14-log([OH^-])=14-log(1.3x10^{-3})=11.11$

That pH allows us to identify the methyl red to perform the titration as long as a pH of about 5, based on the equivalence point, is required.

Best regards.