How many mL (to the nearest mL) of 0.140-M KF solution should be added to 400. mL of 0.212-M HF to prepare a pH

205mL of 0.140M KF solution

Explanation:

pH = 2.70 solution.

It is possible to obtain the pH of the buffer of HF-KF using the H-H equation:

pH = pKa + log [KF] / [HF]

Where pH is desire pH = 2.70

pKa is pKa of HF = 3.17

[KF] could be taken as moles of KF

And [HF] moles of HF: 400.0mL = 0.4L * (0.212mol/L) = 0.0848 moles of HF

Replacing:

2.70 = 3.17 + log [KF] / [0.0848 moles HF]

-0.47 = log [KF] / [0.0848 moles HF]

0.3388 = [KF] / [0.0848 moles HF]

[KF] = 0.02873 moles of KF must be added.

In mL using concentration of KF (0.140M):

0.02873 moles KF * (1L / 0.140 mol) = 0.205L =

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

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