Part A: Are the following solutions acidic, basic, or neutral? [OH?] = 1.0×10 ?7

Question

Part A: Are the following solutions acidic, basic, or neutral?

[OH?] = 1.0×10?7 M

[H3O+] = 3.2×10?3 M

[H3O+] = 9.6×10?4 M

[OH?] = 7.6×10?9 M

Part B: Identify the reactant that is a Brønsted-Lowry acid and the reactant that is a Brønsted-Lowry base in each of the following:

CO32?(aq)+H2O?HCO3?(aq)+OH?(aq) Multiple Choice

NH3(aq)+H2O(l)?NH4+(aq)+OH?(aq) Multiple choice

CH3?COO?(aq)+H3O+(aq)?H2O(l)+CH3?COOH(aq) Multiple choice

Part C: Calculate the [H3O+] of each aqueous solution with the following [OH?]:

NaOH, 6.0×10?3M .  Express your answer using two significant figures.

milk of magnesia, 1.2×10?5M .  Express your answer using two significant figures.

aspirin, 2.4×10?11M . Express your answer using two significant figures.

seawater, 4.0×10?6M . Express your answer using two significant figures.

A. CO32? is the acid (proton acceptor); H2O is the base (proton donor). B. H2O is the acid (proton donor); CO32? is the base (proton acceptor). C. CO32? is the acid (proton donor); H2O is the base (proton acceptor). D. H2O is the acid (proton acceptor); CO32? is the base (proton donor).

Explanation / Answer

A)
[H+] = 1.0*10^-7 M
then, solution is neutral

[H+] > 1.0*10^-7 M
then, acidic

[H+] < 1.0*10^-7 M
then, basic

[OH-] = 1.0*10^-7 M
[H+] = 10^-14/[OH-]
= 10^-14/(1.0*10^-7)
= 1.0*10^-7
neutral solution

[H3O+] = 3.2*10^-3 M
since,
[H3O+] > 1.0*10^-7 M
so, acidic solution

[H3O+] = 9.6*10^-4 M
since,
[H3O+] > 1.0*10^-7 M
so, acidic solution

[OH-] = 7.6*10^-9 M
[H+] = 10^-14/[OH-]
= 10^-14/(7.6*10^-9)
= 1.3*10^-6 M

since, [H+] > 1.0*10^-7 M
so, acidic solution

i am allowed to do only one question at a time

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