11.0.345 moles of NaOH are dissolved in 45.0 mL of water. What is the [H\'] in t

Question

11.0.345 moles of NaOH are dissolved in 45.0 mL of water. What is the [H'] in the solution? a. 7.67 M b. 0.130 M c. 7.67 mM d. 7.67 x 1014 e. 1.30 x 1015 M 12. Which of the following is a basic solution at 25° C? . [H+] # 1.2 x 10.7 b. [OH1-8.3 x 10° c. pOHs 8.3 d. [OH] 1.2 x 10" e, pHs 4.5 13.A solution is prepared by mixing 500.0 mL of 0.10 M NaOCI and 500.0 mL of 0.20 M HOCI. What is the concentration of H' in this solution? For HOcI, K 3.2x 10 a. 7.9 x 10-5 M . 2.6x10"M c. 6.4 x 10-8M d. 1.6 x 10-8 M e. 3.2 x 10-8 M

Explanation / Answer

Given

No. of moles of NaOH = 0.345 moles

Volume = 45 ml = 0.045 L

Molarity of NaOH = 0.345 moles / 0.045 L = 7.67 mol/L

NaOH (aq) ---> Na+ (aq) + OH- (aq)

0.345 moles of NaOH will 0.345 moles of OH-

[OH-] = 0.345 moles / 0.045 L = 7.67 M

generally [H+][OH-] = 10-14

[H+] * 7.67 M = 10-14

[H+] = 1.3 * 10-14 M Answer

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