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3 years ago

9

If the pKa of Lysine is 5, then at a pH of 7, it is most likely going to be a) Neutral. b) Negatively charged. c) Positively cha

rged. d) Partially negatively charged.

1 answer:

Sladkaya [172]3 years ago

7 0

C. Positively charged

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3 years ago

Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C. Solution A: [OH−]=1.55×10−7 M Solution A: [H3O+]= M Solutio

Ahat [919]

Answer: A. $[H_3O^+]=0.64\times 10^{-7}M$

B. $[OH^-]=0.11\times 10^{-5}M$

C. $[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

Thus solution B is basic in nature.

Explanation:

pH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH for acidic solutions is less than 7, for basic solutions it is more than 7 and for neutral solutions it is equal to 7.

$pH=-\log [H^+]$

$pOH=-log[OH^-]$

$pH+pOH=14$

or $[H^+][OH^-]=10^{-14}$

A. $[OH^-]=1.55\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{1.55\times 10^{-7}}=0.64\times 10^{-7}M$

$pH=-log[H_3O^+]=-log[0.64\times 10^{-7}]=7$

B. $[H_3O^+]=9.43\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.43\times 10^{-9}}=0.11\times 10^{-5}M$

$pH=-log[H_3O^+]=-log[9.43\times 10^{-9}]=8$

C. $[H_3O^+]=0.000775M$

$[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

$pH=-log[H_3O^+]=-log[0.000775]=3$

Thus solution B is basic.

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