# If Ca(OH)_2(aq)=0.01*mol*L^-1, what is pH of the solution?

Jul 10, 2016

$p H \cong 12$

#### Explanation:

We know (or should know) that in aqueous solution $p H + p O H = 14$.

Thus, if we calculate $p O H$, $p H$ is directly available.

$p O H = - {\log}_{10} \left(\left[H {O}^{-}\right]\right)$ $= - {\log}_{10} \left(0.02\right)$ (why $0.02$ here and not $0.01$?) $=$ $- \left(- 1.70\right)$ $=$ $1.70$

And from the above, $p H = 14 - 1.70$ $=$ ??
{"ft_lang_label":"__label__en","ft_lang_prob":0.6754582,"math_prob":1.0000091,"size":316,"snap":"2021-31-2021-39","text_gpt3_token_len":91,"char_repetition_ratio":0.10576923,"word_repetition_ratio":0.0,"special_character_ratio":0.278481,"punctuation_ratio":0.10606061,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000073,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-27T04:32:43Z\",\"WARC-Record-ID\":\"<urn:uuid:8bdf7023-6f7e-4975-aa8a-7aa83b0700a6>\",\"Content-Length\":\"32519\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:828e7aab-c8bb-4080-a54a-8981ae9c016d>\",\"WARC-Concurrent-To\":\"<urn:uuid:beeb3106-1f0c-4cfe-a802-9d4b1948a917>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/578212c87c01497b464b70bc\",\"WARC-Payload-Digest\":\"sha1:BODN2M444ROXDYJM42ZHFNTQRL24WYYV\",\"WARC-Block-Digest\":\"sha1:GBGYULVPJYIXHGP2ZWUA4G42ARS4WTEN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046152236.64_warc_CC-MAIN-20210727041254-20210727071254-00492.warc.gz\"}"}