Question

how many milliters of 1.00 m koh should be added to 100 ml of solution containing 10.0 g of histidine hydrochloride (his.hcl fm 191.62) to get a ph of 9.30?.

Answer 1

52.18mL of KOH should be added to 100 mL of solution containing 10.0 g of histidine hydrochloride to get a pH of 9.30.

What is pH?

pH is a measurement of the acidity or basicity of aqueous or other liquid solutions in terms of digits. The phrase, which is frequently used in chemistry, biology, and agronomy, converts hydrogen ion concentrations, which typically vary between 1 and 1014 gram-equivalents per liter, into numbers between 0 and 14.

Given:

pH = 9.30

pH = -log [$H^{+}$]

9.30 = -log [$H^{+}$]

antilog (-9.30) = [$H^{+}$]

[$H^{+}$] = 5.01 × $10^{-10}$ M

Strong acid (His HCl) + strong base (KOH)

[$H^{+}$] = $\frac{-M_{B}V_{B}n_{fB} + M_{A}V_{A}n_{fA}}{V_{final}}$

$M_{B}$ and $M_{A}$= Molarity

$V_{A}$ and $V_{B}$ = Volume

$n_{fA}$ = n factor for acid

$n_{fB}$ = n factor for KOH

5.01 × $10^{-10}$ M = $\frac{-1 M X V_{B} X 1 + 0.52 X 100 X 1 }{V_{A} + V_{B} }$

                       = $\frac{-V_{B} + 52.18 }{100 + V_{B} }$

$n_{fB}$ = $n_{fn}$ = 1

5.01 × $10^{-10}$ M (100 + $V_{B}$ )= $V_{B} - 52.18$

5.01 × $10^{-10}$ + 5.01 × $10^{-10}$ $V_{B}$ = -$V_{B}$ + 52.18

2.18 - 5.01 × $10^{-8}$ = 5.01 × $10^{-10}$ $V_{B}$ + $V_{B}$

$V_{B}$ = 52.17999 mL ≅ 52.18mL of KOH.

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